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Section 4.2 Operations on matrices

In this section we will see some fundamental ways of combining linear transformations to make other linear transformations. As we saw in the previous section, linear transformations are fundamentally related to matrices, so we will also see how the operations we introduce act on matrices.

Subsection 4.2.1 Addition

Definition 4.2.1.

Suppose that \(T : \mathbb{R}^n \to \mathbb{R}^m\) and \(S : \mathbb{R}^n \to \mathbb{R}^m\) are linear transformations. We define their sum to be the function \(T+S : \mathbb{R}^n \to \mathbb{R}^m\) defined by

\begin{equation*} (T+S)(\vec{v}) = T(\vec{v}) + S(\vec{v})\text{.} \end{equation*}

Let \(T : \mathbb{R}^2 \to \mathbb{R}^3\) be defined by \(T\left(\begin{bmatrix}x\\y\end{bmatrix}\right) = \begin{bmatrix}x+y\\x-y\\2y\end{bmatrix}\) and let \(S : \mathbb{R}^2 \to \mathbb{R}^3\) be defined by \(S\left(\begin{bmatrix}x\\y\end{bmatrix}\right) = \begin{bmatrix}y\\x\\2x+y\end{bmatrix}\text{.}\) Then

\begin{equation*} (T+S)\left(\begin{bmatrix}x\\y\end{bmatrix}\right) = T\left(\begin{bmatrix}x\\y\end{bmatrix}\right) + S\left(\begin{bmatrix}x\\y\end{bmatrix}\right) = \begin{bmatrix}x+y\\x-y\\2y\end{bmatrix} + \begin{bmatrix}y\\x\\2x+y\end{bmatrix} = \begin{bmatrix}x+2y \\ 2x-y \\ 2x+3y\end{bmatrix}\text{.} \end{equation*}

Note 4.2.3.

We can only add two linear transformations if they have the same domain (that is, they take inputs from the same dimension as each other) and the same codomain (that is, they give their outputs in the same dimension as each other). For instance, if \(T : \mathbb{R}^2 \to \mathbb{R}^3\) and \(S : \mathbb{R}^3 \to \mathbb{R}^3\) are linear transformations then \(T+S\) does not make sense.

Suppose that \(\vec{v}, \vec{w}\) are vectors in \(\mathbb{R}^n\text{,}\) and \(c\) is a scalar. Then:

\begin{align*} (T+S)(\vec{v}+\vec{w}) \amp = T(\vec{v}+\vec{w}) + S(\vec{v}+\vec{w}) \\ \amp = T(\vec{v}) + T(\vec{w}) + S(\vec{v}) + S(\vec{w}) \\ \amp = T(\vec{v}) + S(\vec{v}) + T(\vec{w}) + S(\vec{w}) \\ \amp = (T+S)(\vec{v}) + (T+S)(\vec{w})\text{,} \end{align*}

and

\begin{align*} (T+S)(c\vec{v}) \amp = T(c\vec{v}) + S(c\vec{v}) \\ \amp = cT(\vec{v}) + cS(\vec{v}) \\ \amp = c(T(\vec{v}) + S(\vec{v})) \\ \amp = c(T+S)(\vec{v}) \text{.} \end{align*}

On the side of matrices, we define addition in the most natural way.

Definition 4.2.5.

Suppose that \(A\) and \(B\) are \(m \times n\)matrices. Their sum is the \(m \times n\) matrix \(A+B\text{,}\) defined by saying that for every \(i\) and \(j\text{,}\) the \((i, j)\) entry of \(A+B\) is the \((i,j)\) entry of \(A\) plus the \((i,j)\) entry of \(B\text{.}\)

If \(A = \begin{bmatrix}1 \amp 2 \amp 3 \\ -2 \amp 0 \amp 1\end{bmatrix}\) and \(B = \begin{bmatrix}0 \amp 1 \amp -1 \\ 4 \amp 1 \amp 0\end{bmatrix}\) then \(A+B = \begin{bmatrix}1 \amp 3 \amp 2 \\ 2 \amp 1 \amp 1\end{bmatrix}\text{.}\)

Note 4.2.7.

We can only add two matrices of the same size. For instance, the expression \(\matr{cc}{1 \amp 2 \\ 3 \amp 4} + \matr{ccc}{1 \amp 2 \amp 3 \\ 2 \amp 4 \amp 5}\) simply does not make sense.

As you probably expect, our definitions of addition of linear transformations and of matrices were designed to interact nicely.

The first column of \([T+S]\) is \((T+S)(\vec{e_1})\text{,}\) and \((T+S)(\vec{e_1}) = T(\vec{e_1}) + S(\vec{e_1})\text{,}\) which is the sum of the first column of \([T]\) and the first column of \([S]\text{.}\) Thus \([T+S]\) and \([T]+[S]\) have the same first column. Repeating this for the other columns, using the other \(\vec{e_k}\)s instead of \(\vec{e_1}\text{,}\) shows that \([T+S]\) and \([T]+[S]\) have exactly the same columns, so \([T+S] = [T] + [S]\text{.}\)

Consider again the transformations \(T : \mathbb{R}^2 \to \mathbb{R}^3\) defined by \(T\left(\begin{bmatrix}x\\y\end{bmatrix}\right) = \begin{bmatrix}x+y\\x-y\\2y\end{bmatrix}\) and \(S : \mathbb{R}^2 \to \mathbb{R}^3\) defined by \(S\left(\begin{bmatrix}x\\y\end{bmatrix}\right) = \begin{bmatrix}y\\x\\2x+y\end{bmatrix}\text{.}\) We saw in Example 4.2.2 that \((T+S) : \mathbb{R}^2 \to \mathbb{R}^3\) is given by \((T+S)\left(\begin{bmatrix}x\\y\end{bmatrix}\right) = \begin{bmatrix}x+2y \\ 2x-y \\ 2x+3y\end{bmatrix}\text{.}\) Plugging in \(\vec{e_1}\) and \(\vec{e_2}\text{,}\) we find that

\begin{equation*} [T+S] = \begin{bmatrix}1 \amp 2 \\ 2 \amp -1 \\ 2 \amp 3\end{bmatrix}\text{.} \end{equation*}

On the other hand, we can also calculate \([T] = \begin{bmatrix}1 \amp 1 \\ 1 \amp -1 \\ 0 \amp 2\end{bmatrix}\) and \([S] = \begin{bmatrix}0 \amp 1 \\ 1 \amp 0 \\ 2 \amp 1\end{bmatrix}\text{,}\) and then calculate

\begin{equation*} [T]+[S] = \begin{bmatrix}1 \amp 2 \\ 2 \amp -1 \\ 2 \amp 3\end{bmatrix}\text{,} \end{equation*}

as we expected.

Definition 4.2.10.

For any \(m\) and \(n\text{,}\) the \(m \times n\) zero matrix is the \(m \times n\) matrix \(0_{m\times n}\) where every entry is \(0\text{.}\)

Subsection 4.2.2 Scalar multiplication

Definition 4.2.11.

Let \(T : \mathbb{R}^n \to \mathbb{R}^m\) be a linear transformation, and let \(c\) be a scalar. We define a new linear transformation \(cT : \mathbb{R}^n \to \mathbb{R}^m\) by

\begin{equation*} (cT)(\vec{v}) = c(T(\vec{v}))\text{.} \end{equation*}

Let \(T : \mathbb{R}^4 \to \mathbb{R}^2\) be defined by \(T\left(\begin{bmatrix}x\\y\\z\\w\end{bmatrix}\right) = \begin{bmatrix}x+y-2w\\z+w\end{bmatrix}\text{.}\) Then \((-3)T : \mathbb{R}^4 \to \mathbb{R}^2\) is defined by

\begin{equation*} ((-3)T)\left(\begin{bmatrix}x\\y\\z\\w\end{bmatrix}\right) = -3\left(T\left(\begin{bmatrix}x\\y\\z\\w\end{bmatrix}\right)\right) = -3\begin{bmatrix}x+y-2w\\z+w\end{bmatrix} = \begin{bmatrix}-3x-3y+6w \\ -3z-3w\end{bmatrix}\text{.} \end{equation*}

This is left as an exercise; it is similar in spirit to the proof of Theorem 4.2.4.

On the side of matrices, we have:

Definition 4.2.14.

Let \(A\) be an \(m \times n\) matrix, and let \(c\) be a scalar. We define the \(m \times n\) matrix \(cA\) by multiplying every entry of \(A\) by \(c\text{.}\)

If \(A = \begin{bmatrix}1 \amp 2 \amp 3 \\ 4 \amp 5 \amp 6\end{bmatrix}\) then \(5A = \begin{bmatrix}5 \amp 10 \amp 15 \\ 20 \amp 25 \amp 30\end{bmatrix}\text{.}\)

Like with addition, the linear transformations viewpoint and the matrix viewpoint play nicely together:

Consider again the transformation \(T : \mathbb{R}^4 \to \mathbb{R}^2\) defined by \(T\left(\begin{bmatrix}x\\y\\z\\w\end{bmatrix}\right) = \begin{bmatrix}x+y-2w\\z+w\end{bmatrix}\text{.}\) We saw in Example 4.2.12 that

\begin{equation*} (-3)T\left(\begin{bmatrix}x\\y\\z\\w\end{bmatrix}\right) = \begin{bmatrix}-3x-3y+6w \\ -3z-3w\end{bmatrix}\text{.} \end{equation*}

A direct calculation using Definition 4.1.8 shows that \([T] = \begin{bmatrix}1 \amp 1 \amp 0 \amp -2 \\ 0 \amp 0 \amp 1 \amp 1\end{bmatrix}\text{,}\) and that

\begin{equation*} (-3)[T] = (-3)\begin{bmatrix}1 \amp 1 \amp 0 \amp -2 \\ 0 \amp 0 \amp 1 \amp 1\end{bmatrix} = \begin{bmatrix}-3 \amp -3 \amp 0 \amp 6 \\ 0 \amp 0 \amp -3 \amp -3\end{bmatrix} = [(-3)T]\text{.} \end{equation*}

As we did with vectors (and also with numbers), we will often write \(A - B\) instead of \(A + (-1)B\text{.}\)

Subsection 4.2.3 Transpose

Our last basic operation is a bit harder to motivate than addition and scalar multiplication, but we will see in Chapter 6 that is is very important. For this particular operation the definition in terms of linear transformations is outside the scope of this course, so we restrict ourselves to matrices.

Definition 4.2.18.

Suppose that \(A\) is an \(m \times n\) matrix. The transpose of \(A\) is the \(n \times m\) matrix obtained by writing the columns of \(A\) as rows.

A matrix \(A\) is called symmetric if \(A = A^t\text{.}\)

Let \(A = \begin{bmatrix}1 \amp 2 \amp 3 \\ 3 \amp 4 \amp 0\end{bmatrix}\text{.}\) Then \(A^t = \begin{bmatrix}1 \amp 3 \\ 2 \amp 4 \\ 3 \amp 0\end{bmatrix}\text{.}\)

If \(A\) is \(m \times n\) then \(A^t\) is \(n \times m\text{,}\) so to have \(A = A^t\) we must have \(m=n\text{,}\) i.e., that \(A\) is square.

Note 4.2.21.

Be careful! Theorem 4.2.20 says that every symmetric matrix is square. It does not say that every square matrix is symmetric. For example, \(\begin{bmatrix}1 \amp 1 \\ 2 \amp 1\end{bmatrix}\) is a square matrix that is not symmetric.

Now that we have several things that we can do to matrices (or linear transformations) it is important to know how these operations interact with each other. Fortunately, the answer is that they interact in exactly the ways that you would hope for, as long as we are careful to make sure that the sizes of the matrices (equivalently, the domains and codomains of the linear transformations) match up correctly. We'll state the relations for matrices, and leave it up to you to translate these statements into the language of linear transformations. We omit the proofs, and invite you to verify these statements yourself.

Exercises 4.2.4 Exercises

1.

Let \(A = \begin{bmatrix} 2 \amp 1 \\ 0 \amp -1 \end{bmatrix}, B = \begin{bmatrix} 3 \amp -1 \amp 2\\ 0 \amp 1 \amp 4 \end{bmatrix}, C = \begin{bmatrix} 3 \amp -1 \\ 2 \amp 0 \end{bmatrix}, D = \begin{bmatrix} 1 \amp 3 \\ -1 \amp 0 \\ 1 \amp 4 \end{bmatrix},\) and \(E = \begin{bmatrix} 1 \amp 0 \amp 1 \\ 0 \amp 1 \amp 0 \end{bmatrix}.\) Compute the following (where possible).

  1. \(\displaystyle 3A - 2B \)

  2. \(\displaystyle 5C \)

  3. \(\displaystyle 3E^t \)

  4. \(\displaystyle B+D \)

  5. \(\displaystyle 4A^t - 3C \)

  6. \(\displaystyle (A+C)^t \)

  7. \(\displaystyle 2B - 3E \)

  8. \(\displaystyle A - D \)

  9. \(\displaystyle (B- 2E)^t \)

Hint 1.
Recall: If \(A\) and \(B\) are matrices of the same size, their sum\(A+B\) is the matrix formed by adding corresponding entries.
Hint 2.
Recall: If \(A\) is any matrix and \(k\) is any number, the scalar multiple\(kA\) is the matrix obtained from \(A\) by multiplying each entry of \(A\) by \(k\) .
Hint 3.
Recall: If \(A\) is a \(m\times n\) matrix, the transpose of \(A\text{,}\) written \(A^t\text{,}\) is the \(n\times m\) matrix whose rows are just the columns of \(A\) in the same order.
Answer.

  1. Answer.

    Does not exist.

  2. Answer.

    \begin{equation*} 5C = \begin{bmatrix} 15 \amp -5 \\ 10 \amp 0 \end{bmatrix}\text{.} \end{equation*}

  3. Answer.

    \begin{equation*} 3E^t = \begin{bmatrix} 3 \amp 0 \\ 0 \amp 3 \\ 3 \amp 0 \end{bmatrix}. \end{equation*}

  4. Answer.

    Does not exist.

  5. Answer.

    \begin{align*} 4A^t - 3C \amp = \begin{bmatrix} -1 \amp 3 \\ -2 \amp -4 \end{bmatrix}. \end{align*}

  6. Answer.

    \begin{align*} (A+C)^t \amp = \begin{bmatrix} 5 \amp 2 \\ 0 \amp -1 \end{bmatrix}. \end{align*}

  7. Answer.

    \begin{align*} 2B - 3E \amp = \begin{bmatrix} 3 \amp -2 \amp 1\\ 0 \amp -1 \amp 8 \end{bmatrix}. \end{align*}

  8. Answer.

    Does not exist.

  9. Answer.

    \begin{align*} (B- 2E)^t \amp = \begin{bmatrix} 1\amp 0 \\ -1 \amp -1 \\ 0 \amp 4 \end{bmatrix}. \end{align*}

Solution.

  1. Solution.

    We cannot add these matrices as they are not of the same size (\(3A\) is a \(2\times 2\) matrix, while \(2B\) is a \(2\times 3\) matrix).

  2. Solution.

    \begin{equation*} 5C = 5\begin{bmatrix} 3 \amp -1 \\ 2 \amp 0 \end{bmatrix} = \begin{bmatrix} 15 \amp -5 \\ 10 \amp 0 \end{bmatrix}\text{.} \end{equation*}

  3. Solution.

    \begin{equation*} 3E^t = 3 \begin{bmatrix} 1 \amp 0 \amp 1 \\ 0 \amp 1 \amp 0 \end{bmatrix}^{T} =3 \begin{bmatrix} 1 \amp 0 \\ 0 \amp 1 \\ 1 \amp 0 \end{bmatrix} = \begin{bmatrix} 3 \amp 0 \\ 0 \amp 3 \\ 3 \amp 0 \end{bmatrix}. \end{equation*}

  4. Solution.

    We cannot add these matrices as they are not of the same size (\(-B\) is a \(2\times 3\) matrix, while \(D\) is a \(3\times 2\) matrix).

  5. Solution.

    \begin{align*} 4A^t - 3C \amp = 4 \begin{bmatrix} 2 \amp 1 \\ 0 \amp -1 \end{bmatrix}^t - 3\begin{bmatrix} 3 \amp -1 \\ 2 \amp 0 \end{bmatrix}\\ \amp = 4 \begin{bmatrix} 2 \amp 0 \\ 1 \amp -1 \end{bmatrix} - \begin{bmatrix} 9 \amp -3 \\ 6 \amp 0 \end{bmatrix}\\ \amp = \begin{bmatrix} 8 - 9 \amp 0 - (-3) \\ 4 - 6 \amp -4 - 0 \end{bmatrix}\\ \amp = \begin{bmatrix} -1 \amp 3 \\ -2 \amp -4 \end{bmatrix}. \end{align*}

  6. Solution.

    \begin{align*} (A+C)^t \amp = \left( \begin{bmatrix} 2 \amp 1 \\ 0 \amp -1 \end{bmatrix} + \begin{bmatrix} 3 \amp -1 \\ 2 \amp 0 \end{bmatrix}\right)^t \\ \amp = \begin{bmatrix} 5 \amp 0 \\ 2 \amp -1 \end{bmatrix}^t \\ \amp = \begin{bmatrix} 5 \amp 2 \\ 0 \amp -1 \end{bmatrix}. \end{align*}

  7. Solution.

    \begin{align*} 2B - 3E \amp = 2 \begin{bmatrix} 3 \amp -1 \amp 2\\ 0 \amp 1 \amp 4 \end{bmatrix} - 3 \begin{bmatrix} 1 \amp 0 \amp 1 \\ 0 \amp 1 \amp 0 \end{bmatrix}\\ \amp = \begin{bmatrix} 6 \amp -2 \amp 4\\ 0 \amp 2 \amp 8 \end{bmatrix} - \begin{bmatrix} 3 \amp 0 \amp 3 \\ 0 \amp 3 \amp 0 \end{bmatrix}\\ \amp = \begin{bmatrix} 3 \amp -2 \amp 1\\ 0 \amp -1 \amp 8 \end{bmatrix}. \end{align*}

  8. Solution.

    We cannot add these matrices as they are not of the same size (\(A\) is a \(2\times 2\) matrix, while \(-D\) is a is a \(3\times 2\) matrix).

  9. Solution.

    \begin{align*} (B- 2E)^t \amp = \left( \begin{bmatrix} 3 \amp -1 \amp 2\\ 0 \amp 1 \amp 4 \end{bmatrix} - 2 \begin{bmatrix} 1 \amp 0 \amp 1 \\ 0 \amp 1 \amp 0 \end{bmatrix}\right)^{T}\\ \amp = \begin{bmatrix} 3 - 2 \amp -1 - 0 \amp 2 - 2\\ 0 - 0 \amp 1 - 2 \amp 4 - 0 \end{bmatrix}^{T}\\ \amp = \begin{bmatrix} 1 \amp -1 \amp 0\\ 0 \amp -1 \amp 4 \end{bmatrix}^{T}\\ \amp = \begin{bmatrix} 1\amp 0 \\ -1 \amp -1 \\ 0 \amp 4 \end{bmatrix}. \end{align*}

2.

Find \(A \) in terms of \(B \) if:

  1. \begin{equation*} A + B = 3A + 2B \end{equation*}

  2. \begin{equation*} 2A - B = 5(A + 2B) \end{equation*}

Hint.
Use the properties from Theorem 4.2.22.
Answer.

  1. \begin{equation*} A= -\frac{1}{2} B \end{equation*}

  2. \begin{equation*} A = -\frac{11}{3}B \end{equation*}

Solution.

Because of the properties listed in Theorem 4.2.22, we can manipulate both sides of the equations just like we would if \(A\) and \(B\) were numbers:

  1. \begin{align*} A + B = 3A + 2B \iff \amp (1-3)A = (2-1)B \\ \iff \amp -2A = B \\ \iff \amp A= -\frac{1}{2} B. \end{align*}

  2. \begin{align*} 2A - B = 5(A + 2B) \iff \amp 2A - B = 5A + 10B\\ \iff \amp -3A = 11B\\ \iff \amp A = -\frac{11}{3}B. \end{align*}

3.

If \(X, Y, A, \) and \(B \) are matrices of the same size, solve the following systems of equations to obtain \(X \) and \(Y \) in terms of \(A \) and \(B \text{.}\)

  1. \begin{align*} 5X + 3Y \amp = A\\ 2X + Y \amp = B \end{align*}

  2. \begin{align*} 4X + 3Y \amp = A\\ 5X + 4Y \amp = B \end{align*}

Hint.
The properties in Theorem 4.2.22 justify manipulating these equations as if \(A,B,X,Y\) were numbers, as long as we do not divide.
Answer.

  1. Answer.

    \(X= -A + 3 B\text{,}\) \(Y=-5B+2A\text{.}\)

  2. Answer.

    \(X= 4A - 3 B\text{,}\) \(Y=4B-5A\)

Solution.

Because of the properties listed in Theorem 4.2.22, we can manipulate both sides of the equations just like we would if \(A, B\) were numbers and \(X, Y\) were variables (however, we cannot simply divide by a matrix!).

  1. Solution.

    The system can also be written as
    \begin{equation*} \matr{cc|c}{ 5 \amp 3 \amp A \\ 2 \amp 1 \amp B}, \end{equation*}
    where it is important that we remember that \(A,B\) are not numbers, so we cannot divide by them. We row-reduce:
    \begin{align*} \matr{cc|c}{5 \amp 3 \amp A \\ 2 \amp 1 \amp B} \underset{5R_{2}}{\overset{2R_{1}}{\longrightarrow}} \amp \matr{cc|c}{10 \amp 6 \amp 2A \\ 10 \amp 5 \amp 5B} \overset{-R_{1}+R_{2}}{\longrightarrow} \matr{cc|c}{10 \amp 6 \amp 2A \\ 0 \amp -1 \amp 5B-2A}\\ \overset{6R_{2}+R_{1}}{\longrightarrow} \amp \matr{cc|c}{10 \amp 0 \amp 2A + 6 (5B-2A) \\ 0 \amp -1 \amp 5B-2A}\\ \amp= \matr{cc|c}{10 \amp 0 \amp -10A + 30 B \\ 0 \amp -1 \amp 5B-2A}. \end{align*}
    We conclude that \(X= \frac{1}{10} (-10A + 30 B) = -A + 3 B\) and \(Y=-5B+2A\text{.}\)

  2. Solution.

    The system can also be written as
    \begin{equation*} \matr{cc|c}{ 4 \amp 3 \amp A \\ 5 \amp 4 \amp B}, \end{equation*}
    where it is important that we remember that \(A,B\) are not numbers, so we cannot divide by them. We row-reduce:
    \begin{align*} \matr{cc|c}{ 4 \amp 3 \amp A \\ 5 \amp 4 \amp B} \overset{5R_{1}}{\underset{4R_{2}}{\longrightarrow}} \amp\matr{cc|c}{ 20 \amp 15 \amp 5A \\ 20 \amp 16 \amp 4B} \overset{-R_{1}+R_{2}}{\longrightarrow} \matr{cc|c}{ 20 \amp 15 \amp 5A \\ 0 \amp 1 \amp 4B-5A}\\ \overset{-15R_{2}+R_{1}}{\longrightarrow} \amp\matr{cc|c}{ 20 \amp 0 \amp 5A-15(4B-5A) \\ 0 \amp 1 \amp 4B-5A}\\ \amp=\matr{cc|c}{ 20 \amp 0 \amp 80A-60B \\ 0 \amp 1 \amp 4B-5A} \end{align*}
    We conclude that \(X= \frac{1}{20} (80A-60B) = 4A - 3 B\) and \(Y=4B-5A\text{.}\)

4.

Let \(A \) and \(B \) be square matrices of the same size. Prove each of the following:

  1. If \(A \) and \(B \) are symmetric, then so is \(A - B \text{.}\)

  2. If \(A\) is symmetric, then so is \(kA \) for any scalar \(k \text{.}\)

Hint.
Recall the definition of symmetric.
Solution.

  1. Solution.

    Let \(A=[a_{ij}]\) and \(B=[b_{ij}]\text{,}\) so that \(A-B=[a_{ij}-b_{ij}]\text{.}\) Since \(A=At\text{,}\) we have \(a_{ij} = a_{ji}\) for all choices of \(i,j\text{,}\) and similarly \(b_{ij} = b_{ji}\text{.}\) Thus,
    \begin{align*} (A-B)^{T} \amp= [a_{ij}-b_{ij}]^{T} = [a_{ji}-b_{ji}] \\ \amp= [a_{ij}-b_{ij}] = [a_{ij}]-[b_{ij}] = A-B, \end{align*}
    so \(A-B\) is symmetric.

  2. Solution.

    We need to show that \((kA)^{T} = k A\) to conclude that \(kA\) is symmetric. Since equality of matrices is entrywise equality, we will consider their \((i,j)\)-entries:

    Let \(A=[a_{ij}]\text{.}\) By definition of scalar multiplication, \(kA=[ka_{ij}]\text{.}\) By definition of the transpose, the \((i,j)\)-entry of \((kA)^{T}\) is therefore \(ka_{ji}\) (where we swapped the roles of \(i\) and \(j\)).

    Since \(A\) is symmetric, we know that \(a_{ij}=a_{ji}\) for any choice of \(i,j\text{,}\) so in particular \(ka_{ij}=ka_{ji}\) also. This means that the \((i,j)\)-entry of \(kA\) (namely \(ka_{ij}\)) coincides with the \((i,j)\)-entry of \((kA)^{T}\) (namely \(ka_{ji}\)), which shows that \(k A = (kA)^{T} \) since they are equal entrywise.

5.

A square matrix is called skew-symmetric if \(A^t = -A \text{.}\) Show that the main diagonal of a skew-symmetric matrix consists of only zeros.
Hint.
If \(A=[a_{ij}]\text{,}\) then the main diagonal consists of every entry of the matrix which is of the form \(a_{ii} \text{.}\)
Solution.

Write \(A=[a_{ij}]\text{.}\) By assumption,
\begin{equation*} [a_{ji}] = A^{t} = -A = [-a_{ij}], \end{equation*}
so that \(a_{ji}=-a_{ij} \) for all choices of \(i,j\text{.}\) In particular, if \(i=j\text{,}\) then \(a_{ii}=-a_{ii} \text{,}\) so \(a_{ii}=0\text{.}\)

6.

Show that \(A + A^t \) is symmetric for any square matrix \(A \text{.}\)
Hint.
Recall the definition of symmetric.
Solution.

Recall first that \((A+B)^t = A^t + B^t\) for any matrices \(A, B\) of the same size. Now, since \(A\) is square, \(A\) and \(A^t\) are of the same size, so we may use this fact for \(B=A^t\text{:}\)
\begin{align*} (A+A^t)^t \amp= A^t + (A^t)^t\\ \amp = A^t + A,\qquad\text{ since } (A^t)^t=A\\ \amp = A + A^t,\qquad\text{ since addition is commutative.} \end{align*}
This shows that \(A+A^t\) is its own transpose, i.e. symmetric.