Calculate the following expressions, using the following matrices:
\[A = \begin{bmatrix} 2 & 3 \\ -5 & 1 \end{bmatrix}, \qquad B = \begin{bmatrix} 0 & -1 \\ 1 & -1 \end{bmatrix}, \qquad \mathbf{v} = \begin{bmatrix} 2 \\ -4 \end{bmatrix}\]

(a) $A B^\trans + \mathbf{v} \mathbf{v}^\trans$.

(b) $A \mathbf{v} – 2 \mathbf{v}$.

(c) $\mathbf{v}^{\trans} B$.

(d) $\mathbf{v}^\trans \mathbf{v} + \mathbf{v}^\trans B A^\trans \mathbf{v}$.

Solution.

(a) $A B^\trans + \mathbf{v} \mathbf{v}^\trans$

\begin{align*} A B^\trans + \mathbf{v} \mathbf{v}^\trans &= \begin{bmatrix} 2 & 3 \\ -5 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -1 & -1 \end{bmatrix} + \begin{bmatrix} 2 \\ -4 \end{bmatrix} \begin{bmatrix} 2 & -4 \end{bmatrix} \\[6pt] &= \begin{bmatrix} -3 & -1 \\ -1 & -6 \end{bmatrix} + \begin{bmatrix} 4 & -8 \\ -8 & 16 \end{bmatrix} \\[6pt] &= \begin{bmatrix} 1 & -9 \\ -9 & 10 \end{bmatrix} \end{align*}

(b) $A \mathbf{v} – 2 \mathbf{v} $

\begin{align*} A \mathbf{v} – 2 \mathbf{v} &= \begin{bmatrix} 2 & 3 \\ -5 & 1 \end{bmatrix} \begin{bmatrix} 2 \\ -4 \end{bmatrix} – 2 \begin{bmatrix} 2 \\ -4 \end{bmatrix} \\[6pt] &= \begin{bmatrix} -8 \\ -14 \end{bmatrix} + \begin{bmatrix} -4 \\ 8 \end{bmatrix} \\[6pt] &= \begin{bmatrix} -12 \\ -6 \end{bmatrix} \end{align*}

(c) $\mathbf{v}^{\trans} B$

\begin{align*} \mathbf{v}^{\trans} B &= \begin{bmatrix} 2 & -4 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & -1 \end{bmatrix} \\[6pt] &= \begin{bmatrix} -4 & 2 \end{bmatrix} \end{align*}

(d) $\mathbf{v}^\trans \mathbf{v} + \mathbf{v}^\trans B A^\trans \mathbf{v}$

\begin{align*}
&\mathbf{v}^\trans \mathbf{v} + \mathbf{v}^\trans B A^\trans \mathbf{v}\\[6pt] &= \begin{bmatrix} 2 & -4 \end{bmatrix} \begin{bmatrix} 2 \\ -4 \end{bmatrix} + \left( \begin{bmatrix} 2 & -4 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & -1 \end{bmatrix} \right) \left( \begin{bmatrix} 2 & -5 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} 2 \\ -4 \end{bmatrix} \right) \\[6pt] &= 20 + \begin{bmatrix} -4 & 2 \end{bmatrix} \begin{bmatrix} 24 \\ 2 \end{bmatrix} \\[6pt] &= 20 – 92 \\
&= -72 \end{align*}

The post Matrix Operations with Transpose first appeared on Problems in Mathematics.

A square matrix $A$ is called idempotent if $A^2=A$.

(a) Suppose $A$ is an $n \times n$ idempotent matrix and let $I$ be the $n\times n$ identity matrix. Prove that the matrix $I-A$ is an idempotent matrix.

(b) Assume that $A$ is an $n\times n$ nonzero idempotent matrix. Then determine all integers $k$ such that the matrix $I-kA$ is idempotent.

(c) Let $A$ and $B$ be $n\times n$ matrices satisfying
\[AB=A \text{ and } BA=B.\] Then prove that $A$ is an idempotent matrix.

Proof.

(a) Prove that the matrix $I-A$ is an idempotent matrix.

To prove that $I-A$ is an idempotent matrix, we show that $(I-A)^2=I-A$.
We compute
\begin{align*}
(I-A)^2&=(I-A)(I-A)\\
&=I(I-A)-A(I-A)\\
&=I-A-A+A^2\\
&=I-2A+A^2\\
&=I-2A+A && \text{since $A$ is idempotent}\\
&=I-A.
\end{align*}
Thus, we have $(I-A)^2=I-A$ and $I-A$ is an idempotent matrix.

(b) Determine all integers $k$ such that the matrix $I-kA$ is idempotent.

Let us find out the condition on $k$ so that $I-kA$ is an idempotent matrix.
We have
\begin{align*}
&(I-kA)^2=(I-kA)(I-kA)\\
&=I(I-kA)-kA(I-kA)\\
&=I-kA-kA+k^2A^2\\
&=I-2kA+k^2A && \text{ since $A$ is idempotent}\\
&=I-(2k-k^2)A.
\end{align*}

It follows that $I-kA$ is idempotent if and only if $I-kA=I-(2k-k^2)A$, or equivalently $(k^2-k)A=O$, the zero matrix.
Since $A$ is not the zero matrix, we see that $I-kI$ is idempotent if and only if $k^2-k=0$.

Since $k^2-k=k(k-1)$, we conclude that $I-kA$ is an idempotent matrix if and only if $k=0, 1$.

(c) Prove that $A$ is an idempotent matrix.

Let $A$ and $B$ be $n\times n$ matrices satisfying
\[AB=A \tag{*}\] and
\[ BA=B. \tag{**}\] Our goal is to show that $A^2=A$.
We compute
\begin{align*}
&A^2=AA\\
&=(AB)A && \text{by (*)}\\
&=A(BA)\\
&=AB && \text{by (**)}\\
&=A && \text{by (*)}.
\end{align*}
Therefore, we have obtained the identity $A^2=A$, and we conclude that $A$ is an idempotent matrix.

Related Question.

The proofs are given in the post ↴
Unit Vectors and Idempotent Matrices

The post If $A$ is an Idempotent Matrix, then When $I-kA$ is an Idempotent Matrix? first appeared on Problems in Mathematics.