Let $A$ and $B$ be $n \times n$ matrices, and $\mathbf{v}$ an $n \times 1$ column vector.

Use the matrix components to prove that $(A + B) \mathbf{v} = A\mathbf{v} + B\mathbf{v}$.

Solution.

We will use the matrix components $A = (a_{i j})_{1 \leq i, j \leq n}$, $B = (b_{i j})_{1 \leq i , j \leq n }$, and $\mathbf{v} = (v_i)_{1 \leq i \leq n}$.

Then
\begin{align*}
(A + B) \mathbf{v} = \begin{bmatrix} \sum_{j=1}^n (a_{1 j} + b_{1 j} ) v_j \\[3pt] \sum_{j=1}^n (a_{2 j} + b_{2 j} ) v_j \\[3pt]\sum_{j=1}^n (a_{3 j} + b_{3 j} ) v_j \\ \vdots \\
\sum_{j=1}^n (a_{n j} + b_{n j} ) v_j \end{bmatrix}.
\end{align*}

On the other hand,
\[A\mathbf{v} + B\mathbf{v} = \begin{bmatrix} \sum_{j=1}^n a_{1 j} v_j \\[3pt] \sum_{j=1}^n a_{2 j} v_j \\[3pt] \sum_{j=1}^n a_{3 j} v_j \\ \vdots \\ \sum_{j=1}^n a_{n j} v_j \end{bmatrix} + \begin{bmatrix} \sum_{j=1}^n b_{1 j} v_j \\[3pt] \sum_{j=1}^n b_{2 j} v_j \\[3pt] \sum_{j=1}^n b_{3 j} v_j \\
\vdots \\ \sum_{j=1}^n b_{n j} v_j \end{bmatrix} = \begin{bmatrix} \sum_{j=1}^n (a_{1 j} + b_{1 j} ) v_j \\[3pt] \sum_{j=1}^n (a_{2 j} + b_{2 j} ) v_j \\[3pt] \sum_{j=1}^n (a_{3 j} + b_{3 j} ) v_j \\ \vdots \\ \sum_{j=1}^n (a_{n j} + b_{n j} ) v_j \end{bmatrix}.\]

We can see that they are the same.

The post Prove that $(A + B) \mathbf{v} = A\mathbf{v} + B\mathbf{v}$ Using the Matrix Components first appeared on Problems in Mathematics.

Let
\[D=\begin{bmatrix}
d_1 & 0 & \dots & 0 \\
0 &d_2 & \dots & 0 \\
\vdots & & \ddots & \vdots \\
0 & 0 & \dots & d_n
\end{bmatrix}\] be a diagonal matrix with distinct diagonal entries: $d_i\neq d_j$ if $i\neq j$.
Let $A=(a_{ij})$ be an $n\times n$ matrix such that $A$ commutes with $D$, that is,
\[AD=DA.\] Then prove that $A$ is a diagonal matrix.

Proof.

We prove that the $(i,j)$-entry of $A$ is $a_{ij}=0$ for $i\neq j$.

We compare the $(i,j)$-entries of both sides of $AD=DA$.
Let $D=(d_{ij})$. That is, $d_{ii}=d_i$ and $d_{ij}=0$ if $i\neq j$.
The $(i,j)$-entry of $AD$ is
\begin{align*}
(AD)_{ij}=\sum_{k=1}^n a_{ik}d_{kj}=a_{ij}d_j.
\end{align*}

The $(i,j)$-entry of $DA$ is
\[(DA)_{ij}=\sum_{k=1}^n d_{ik}a_{kj}=d_ia_{ij}.\]

Hence we have
\[a_{ij}d_j=a_{ij}d_i.\] Or equivalently we have
\[a_{ij}(d_j-d_i)=0.\]

Since $d_i\neq d_j$, we have $d_j-d_i\neq 0$.
Thus, we must have $a_{ij}=0$ for $i\neq j$.

Hence $A=(a_{ij})$ is a diagonal matrix.

The post A Matrix Commuting With a Diagonal Matrix with Distinct Entries is Diagonal first appeared on Problems in Mathematics.

Let $A$ be the coefficient matrix of the system of linear equations
\begin{align*}
-x_1-2x_2&=1\\
2x_1+3x_2&=-1.
\end{align*}

(a) Solve the system by finding the inverse matrix $A^{-1}$.

(b) Let $\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}$ be the solution of the system obtained in part (a).
Calculate and simplify
\[A^{2017}\mathbf{x}.\]

(The Ohio State University, Linear Algebra Midterm Exam Problem)
 

Solution.

(a) Solve the system by finding the inverse matrix $A^{-1}$.

The coefficient matrix $A$ of the system is
\[A=\begin{bmatrix}
-1 & -2\\
2& 3
\end{bmatrix}.\] The determinant of the matrix $A$ is $\det(A)=(-1)(3)-(-2)(2)=1\neq 0$, and thus $A$ is invertible and the inverse matrix can be found by the $2\times 2$ inverse matrix formula
\[A^{-1}=\frac{1}{ad-bc}\begin{bmatrix}
d & -b\\
-c& a
\end{bmatrix}\] if $A=\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}$ and $\det(A)=ad-bc\neq 0$.
Thus
\[A^{-1}=\begin{bmatrix}
3 & 2\\
-2& -1
\end{bmatrix}.\] (Or you could form the augmented matrix $[A\mid I]$ and find $A^{-1}$.)

Now the given system of linear equations can be written as
\[A\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}=\begin{bmatrix}
1 \\
-1
\end{bmatrix}.\] Multiplying by $A^{-1}$ on the left, we obtain
\[\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}=A^{-1}\begin{bmatrix}
1 \\
-1
\end{bmatrix}=\begin{bmatrix}
3 & 2\\
-2& -1
\end{bmatrix}\begin{bmatrix}
1 \\
-1
\end{bmatrix}=\begin{bmatrix}
1 \\
-1
\end{bmatrix}.\] Thus the solution of the system is $x_1=1, x_2=-1$, or in the vector form
\[\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}=\begin{bmatrix}
1 \\
-1
\end{bmatrix}.\]

(b) Calculate and simplify $A^{2017}\mathbf{x}$

Since $\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}=\begin{bmatrix}
1 \\
-1
\end{bmatrix}$ is the solution of the system $A\mathbf{x}=\begin{bmatrix}
1 \\
-1
\end{bmatrix}$, we see that
\[A\mathbf{x}=\mathbf{x}.\] From this, we have
\begin{align*}
A^2\mathbf{x}&=AA\mathbf{x}=A\mathbf{x}=\mathbf{x}\\
A^3\mathbf{x}&=AA^2\mathbf{x}=A\mathbf{x}=\mathbf{x}\\
A^4\mathbf{x}&=AA^3\mathbf{x}=A\mathbf{x}=\mathbf{x}\\
\vdots &
\end{align*}

Repeating this, we see that
\[A^{2017}\mathbf{x}=\mathbf{x}.\] (To be more precise, you can prove that $A^n\mathbf{x}=\mathbf{x}$ for any positive integer $n$ by mathematical induction on $n$.)
Thus, we have
\[A^{2017}\mathbf{x}=\begin{bmatrix}
1 \\
-1
\end{bmatrix}.\]

Comment.

This is one of the first midterm exam problems of linear algebra (Math 2568) at the Ohio State University.

Midterm 1 problems and solutions

The following list is the problems and solutions/proofs of midterm exam 1 of linear algebra at the Ohio State University in Spring 2017.

1. Problem 1 and its solution: Possibilities for the solution set of a system of linear equations
2. Problem 2 and its solution: The vector form of the general solution of a system
3. Problem 3 and its solution: Matrix operations (transpose and inverse matrices)
4. Problem 4 and its solution: Linear combination
5. Problem 5 and its solution: Inverse matrix
6. Problem 6 and its solution: Nonsingular matrix satisfying a relation
7. Problem 7 and its solution (The current page): Solve a system by the inverse matrix
8. Problem 8 and its solution:A proof problem about nonsingular matrix

The post Solve a System by the Inverse Matrix and Compute $A^{2017}\mathbf{x}$ first appeared on Problems in Mathematics.

Let $A, B, C$ be the following $3\times 3$ matrices.
\[A=\begin{bmatrix}
1 & 2 & 3 \\
4 &5 &6 \\
7 & 8 & 9
\end{bmatrix}, B=\begin{bmatrix}
1 & 0 & 1 \\
0 &3 &0 \\
1 & 0 & 5
\end{bmatrix}, C=\begin{bmatrix}
-1 & 0\ & 1 \\
0 &5 &6 \\
3 & 0 & 1
\end{bmatrix}.\] Then compute and simplify the following expression.
\[(A^{\trans}-B)^{\trans}+C(B^{-1}C)^{-1}.\]

(The Ohio State University, Linear Algebra Midterm Exam Problem)
 

Solution.

We first use the properties of transpose matrices and inverse matrices and simplify the expression.
Note that we have
\[(A^{\trans}-B)^{\trans}=(A^{\trans})^{\trans}-B^{\trans}=A-B\] since the double transpose $(A^{\trans})^{\trans}=A$ and $B$ is a symmetric matrix.

Also, note that we have
\[(B^{-1}C)^{-1}=C^{-1}(B^{-1})^{-1}=C^{-1}B\] since $(B^{-1})^{-1}=B$. Care must be taken when you distribute the inverse sign in the first equality. We needed to switch the order of the product.
Then we have
\[C(B^{-1}C)^{-1}=CC^{-1}B=IB=B,\] where $I$ is the $3\times 3$ identity matrix.

Therefore, the given expression can be simplified into
\[(A^{\trans}-B)^{\trans}+C(B^{-1}C)^{-1}=A-B+B=A.\] Hence we have
\[(A^{\trans}-B)^{\trans}+C(B^{-1}C)^{-1}=A=\begin{bmatrix}
1 & 2 & 3 \\
4 &5 &6 \\
7 & 8 & 9
\end{bmatrix}.\]

Comment.

This is one of the midterm exam 1 problems of linear algebra (Math 2568) at the Ohio State University.

Common Mistakes

Here are two common mistakes.

The first one is
\[(B^{-1}C)^{-1}=(B^{-1})^{-1}C^{-1} \quad \text{(This is wrong!!)}.\] Note that in general, we have
\[(AB)^{-1}=B^{-1}A^{-1} \quad \text{(Note the order of products)}.\]

The second common mistake is that
\[CBC^{-1}=CC^{-1}B=B \quad \text{(The first equality is wrong!!)}.\] Recall that in general matrix multiplication is not commutative, meaning that
\[AB\neq BA.\]

And surprisingly, if you combine these two mistakes, you get the right answer.
\[C(B^{-1}C)^{-1}=CBC^{-1}=B. \quad \text{(This is wrong!!)}.\]

However, these mistakes show that you didn’t understand matrix operations including transpose and inverse matrices.

Midterm 1 problems and solutions

The following list is the problems and solutions/proofs of midterm exam 1 of linear algebra at the Ohio State University in Spring 2017.

1. Problem 1 and its solution: Possibilities for the solution set of a system of linear equations
2. Problem 2 and its solution: The vector form of the general solution of a system
3. Problem 3 and its solution (The current page): Matrix operations (transpose and inverse matrices)
4. Problem 4 and its solution: Linear combination
5. Problem 5 and its solution: Inverse matrix
6. Problem 6 and its solution: Nonsingular matrix satisfying a relation
7. Problem 7 and its solution: Solve a system by the inverse matrix
8. Problem 8 and its solution:A proof problem about nonsingular matrix

The post Compute and Simplify the Matrix Expression Including Transpose and Inverse Matrices first appeared on Problems in Mathematics.

Let $A$ and $B$ are $n \times n$ matrices with real entries.
Assume that $A+B$ is invertible. Then show that
\[A(A+B)^{-1}B=B(A+B)^{-1}A.\]

(University of California, Berkeley Qualifying Exam)

Proof.

Let $P=A+B$. Then $B=P-A$.

Using these, we express the given expression in terms of only $A$ and $P$.
On one hand, we have
\[A(A+B)^{-1}B=AP^{-1}(P-A)=AP^{-1}P-AP^{-1}A=A-AP^{-1}A.\] On the other hand we have
\[B(A+B)^{-1}A=(P-A)P^{-1}A=PP^{-1}A-AP^{-1}A=A-AP^{-1}A.\] Thus these are equal.

This completes the proof.

Comment.

Did I make a mistake? Isn’t it too simple to be a qualifying exam problem?

The post Simple Commutative Relation on Matrices first appeared on Problems in Mathematics.