Linear Algebra - The Transpose

True or false: The transpose of a matrix with six rows and seven columns has seven rows and six columns.

True

False

Explanation

The transposition of a matrix switches the rows and the columns, so the number of rows in the original matrix is equal to the number of columns in the transpose, and vice versa. Therefore, a matrix with six rows and seven columns has seven rows and six columns.

$\begin{align*}&\text{Find the transpose of }A=\begin{bmatrix}-15&3\-2&18\end{bmatrix}\end{align*}$

$\begin{bmatrix}-15&-2\3&18\end{bmatrix}$

$\begin{bmatrix}3&18\-15&-2\end{bmatrix}$

$\begin{bmatrix}18&3\-2&-15\end{bmatrix}$

$\begin{bmatrix}-2&-15\18&3\end{bmatrix}$

Explanation

$\begin{align*}&\text{Determining the transpose of a matrix is relatively simple, though attention to detail should be exercised.}\&\text{In creating the transpose, the first row becomes the first column, the second row the second column, etc:}\&A=\begin{bmatrix}-15&3\-2&18\end{bmatrix}\&A^{T}=\begin{bmatrix}-15&-2\3&18\end{bmatrix}\end{align*}$

Find the transpose of Matrix .

$A=\begin{bmatrix} 1& 4\ 5& 8 \end{bmatrix}$

$A^T=\begin{bmatrix} 1& 5\ 4& 8 \end{bmatrix}$

$A^T=\begin{bmatrix} 1& 4\ 5& 8 \end{bmatrix}$

$A^T=\begin{bmatrix} 1& 5\ 8& 4 \end{bmatrix}$

$A^T=\begin{bmatrix} 8& 5\ 4& 1 \end{bmatrix}$

$A^T=\begin{bmatrix} 4& 5\ 8& 1 \end{bmatrix}$

Explanation

To find the transpose, we need to make columns into rows.

$A^T=\begin{bmatrix} 1& 5\ 4& 8 \end{bmatrix}$

is a two-by-two matrix such that

$A + A^{T} = \begin{bmatrix} -10& 0\ 0& 16 \end{bmatrix}$

Which of the following could be?

$A = \begin{bmatrix} -5 &- 11 \ 11 & 8 \end{bmatrix}$

$A = \begin{bmatrix} -5 &12 \ 12& 8 \end{bmatrix}$

$A = \begin{bmatrix} -5 & - 7 \ -7& 8 \end{bmatrix}$

$A = \begin{bmatrix} -10 &- 11 \ 11 & 16 \end{bmatrix}$

$A = \begin{bmatrix} -10 &- 7\ -7& 16 \end{bmatrix}$

Explanation

Since is a two-by-two matrix, we will let

$A = \begin{bmatrix} a& b\ c& d \end{bmatrix}$ ,

where , , , and represent real values.

The transpose of a matrix switches the rows and the columns. Therefore, the first column of $A^{T}$ has the same entries, in order, as the first row of , and so forth. Therefore,

$A ^{T}= \begin{bmatrix} a&c\ b& d \end{bmatrix}$

and

$A+A^{T} = \begin{bmatrix} a& b\ c& d \end{bmatrix} + \begin{bmatrix} a&c\ b& d \end{bmatrix}$

Add two matrices termwise:

$A+A^{T} = \begin{bmatrix} a+a & b+c\ c+b& d+d \end{bmatrix}$

$= \begin{bmatrix} 2a & b+c\ c+b& 2d \end{bmatrix}$

Since

$A + A^{T} = \begin{bmatrix} -10& 0\ 0& 16 \end{bmatrix}$

or, equivalently,

$\begin{bmatrix} 2a & b+c\ c+b& 2d \end{bmatrix}= \begin{bmatrix} -10& 0\ 0& 16 \end{bmatrix}$

$\begin{bmatrix} 2a & b+c\ b+c& 2d \end{bmatrix}= \begin{bmatrix} -10& 0\ 0& 16 \end{bmatrix}$

The following must hold:

and

or, equivalently,

Therefore, any matrix of the form

$A = \begin{bmatrix} -5 & b\ -b & 8 \end{bmatrix}$

for some real would make the statement correct. The only choice matching this pattern is the matrix

$A = \begin{bmatrix} -5 &- 11 \ 11 & 8 \end{bmatrix}$ .

$\begin{align*}&\text{Find the transpose of }\begin{bmatrix}1&2\-2&-6\-8&-20\-18&-14\16&15\20&3\end{bmatrix}^{T}\end{align*}$

$\begin{bmatrix}1&2\-2&-6\-8&-20\-18&-14\16&15\20&3\end{bmatrix}$

$\begin{bmatrix}1&-2&-8&-18&16&20\2&-6&-20&-14&15&3\end{bmatrix}$

$\begin{bmatrix}3&15&-14&-20&-6&2\20&16&-18&-8&-2&1\end{bmatrix}$

$\begin{bmatrix}20&16&-18&-8&-2&1\3&15&-14&-20&-6&2\end{bmatrix}$

Explanation

$\begin{align*}&\text{Note the superscript, T, in our problem statement; this denotes the transpose.}\&\text{The transpose of the transpose of a matrix is the orignal matrix, so:}\&\begin{bmatrix}1&2\-2&-6\-8&-20\-18&-14\16&15\20&3\end{bmatrix}^{TT}=\begin{bmatrix}1&2\-2&-6\-8&-20\-18&-14\16&15\20&3\end{bmatrix}\end{align*}$

Let and be any two matrices of the same dimensions.

True or false:

It must hold that $(A+B)^{T} = A^{T}+ B^{T}$ .

True

False

Explanation

The transpose of the sum of two matrices is indeed equal to the sum of their transposes. Let us look at the two-by-two case - this reasoning can be generalized.

$A = \begin{bmatrix} a& b\ c& d \end{bmatrix}$ and $B = \begin{bmatrix} w & x\ y & z \end{bmatrix}$

The transpose of a matrix switches the rows and the columns. Therefore, the first column of $A^{T}$ has the same entries, in order, as the first row of , and so forth. Therefore,

$A^{T} = \begin{bmatrix} a& c\ b& d \end{bmatrix}$ and $B ^{T}= \begin{bmatrix} w & y\x& z \end{bmatrix}$ .

Matrices are added term by term, so

$A^{T}+ B ^{T}$

$= \begin{bmatrix} a& c\ b& d \end{bmatrix} + \begin{bmatrix} w & y\x& z \end{bmatrix}$

$= \begin{bmatrix} a+w & c+y\ b+x& d+z \end{bmatrix}$

$(A+B)^{T}$ , the transpose of the sum of the original matrices, can be found by first adding the matrices termwise:

$= \begin{bmatrix} a& b\ c& d \end{bmatrix}+ \begin{bmatrix} w & x\ y & z \end{bmatrix}$

$= \begin{bmatrix} a+w &b+x \ c+y & d+z \end{bmatrix}$

Take the transpose of the sum:

$(A+B)^{T}= \begin{bmatrix} a+w & c+y\ b+x& d+z \end{bmatrix}$

Indeed, $(A+B)^{T} = A^{T}+ B^{T}$ .

$A = \begin{bmatrix} 3 & 0 & -3 \ 6& 0 & 9 \end{bmatrix}$

Which of the following is equal to $A^{T}$ ?

$A ^{T}= \begin{bmatrix} 3 &6 \ 0 & 0 \ -3 & 9 \end{bmatrix}$

$A ^{T}= \begin{bmatrix}6 & 3 \ 0 & 0 \ -9 & 3 \end{bmatrix}$

$A^{T} = \begin{bmatrix} 6& 0 & 9 \ 3 & 0 & -3 \end{bmatrix}$

$A^{T} = \begin{bmatrix} -3 & 0 & 3 \ -6& 0 &- 9 \end{bmatrix}$

The transpose of does not exist.

Explanation

The transpose of a matrix switches the rows and the columns. Therefore, the first column of $A^{T}$ has the same entries, in order, as the first row of , and so forth. Since

$A = \begin{bmatrix} 3 & 0 & -3 \ 6& 0 & 9 \end{bmatrix}$

it follows that

$A ^{T}= \begin{bmatrix} 3 &6 \ 0 & 0 \ -3 & 9 \end{bmatrix}$

Which is the transpose of $\begin{bmatrix} 1 &8 & 4\2 &-3 &0 \end{bmatrix}$ ?

$\begin{bmatrix} 1 &2 \8 & -3\ 4& 0\end{bmatrix}$

$\begin{bmatrix} 2 & 1\ -3& 8\ 0& 4\end{bmatrix}$

$\begin{bmatrix} 1 &4 \2 &-3 \8 &0 \end{bmatrix}$

$\begin{bmatrix} 1 & 2& 8\ 4& -3& 0\end{bmatrix}$

Explanation

The transverse is the matrix where the columns are now the corresponding rows - the first column is now the first row, the second column is now the second row, etc.

$A^{*} = \begin{bmatrix} 3 + i & 5 - 2i \3 + 2i & 3i \end{bmatrix}$

Which of the following is equal to $A^{T}$ ?

$A^{T} = \begin{bmatrix} 3 - i & 5 + 2i \3 - 2i & -3i \end{bmatrix}$

$A^{T} = \begin{bmatrix} 3 - i & 5 + 2i \3 - 2i & 3i \end{bmatrix}$

$A^{T} = \begin{bmatrix} 3 + i & 5 + 2i \3 - 2i & 3i \end{bmatrix}$

$A^{T} = \begin{bmatrix} 3 - i &3 - 2i \ 5 + 2i & -3i \end{bmatrix}$

$A^{T} = \begin{bmatrix} 3 + i &3 - 2i \ 5 + 2i & 3i \end{bmatrix}$

Explanation

$A^{T}$ is the transpose of - the result of interchanging the rows of with its columns. $A^{*}$ is the conjugate transpose of - the result of changing each entry of $A^{T}$ to its complex conjugate. Therefore, if