The following problems are True or False.

Let $A$ and $B$ be $n\times n$ matrices.

(a) If $AB=B$, then $B$ is the identity matrix.
(b) If the coefficient matrix $A$ of the system $A\mathbf{x}=\mathbf{b}$ is invertible, then the system has infinitely many solutions.
(c) If $A$ is invertible, then $ABA^{-1}=B$.
(d) If $A$ is an idempotent nonsingular matrix, then $A$ must be the identity matrix.
(e) If $x_1=0, x_2=0, x_3=1$ is a solution to a homogeneous system of linear equation, then the system has infinitely many solutions.

Solution.

(a) True or False: if $AB=B$, then $B$ is the identity matrix.

False. For example, if $B$ is the zero matrix, then of course we have $AB=B$ as both sides are the zero matrix.

(b) True or False: if the coefficient matrix $A$ of the system $A\mathbf{x}=\mathbf{b}$ is invertible, then the system has infinitely many solutions.

False. If the coefficient matrix $A$ is invertible, the system has a unique solution $\mathbf{x}=A^{-1}\mathbf{b}$.

(c) True or False: if $A$ is invertible, then $ABA^{-1}=B$.

False. The given equality is equivalent to $AB=BA$. Even $A$ is invertible, matrix multiplication is not commutative. As a counterexample, consider
\[A=\begin{bmatrix}
1 & 1\\
0& 1
\end{bmatrix} \text{ and } B=\begin{bmatrix}
1 & 0\\
-1& 1
\end{bmatrix}.\] Note that the determinant of $A$ is $\det(A)=1\neq 0$. Hence $A$ is invertible.
Yet, we have
\[AB=\begin{bmatrix}
1 & 1\\
0& 1
\end{bmatrix}\begin{bmatrix}
1 & 0\\
-1& 1
\end{bmatrix}=\begin{bmatrix}
0 & 1\\
-1& 1
\end{bmatrix}\] and
\[BA=\begin{bmatrix}
1 & 0\\
-1& 1
\end{bmatrix}\begin{bmatrix}
1 & 1\\
0& 1
\end{bmatrix}=\begin{bmatrix}
1 & 1\\
-1& 0
\end{bmatrix},\] and hence $AB\neq BA$.

(d) True or False: if $A$ is an idempotent nonsingular matrix, then $A$ must be the identity matrix.

True. Since $A$ is idempotent, we have $A^2=A$. As $A$ is nonsingular, it is invertible. Thus, the inverse matrix $A^{-1}$ exists. Then we have
\[I=A^{-1}A=A^{-1}A^2=IA=A.\] Hence, such a matrix $A$ must be the identity matrix $I$.

(e) True or False: if $x_1=0, x_2=0, x_3=1$ is a solution to a homogeneous system of linear equation, then the system has infinitely many solutions.

Note that any homogeneous system has the zero solution. In addition to the zero solution, this system has a solution $x_1=0, x_2=0, x_3=1$. So, it has at least two solutions. The only possibilities for the number of solutions of a system of linear equations are zero, one, or infinitely many.
So, we conclude that the system must have infinitely many solutions.

Let $A$ be an $n\times n$ nonsingular matrix. Let $\mathbf{v}, \mathbf{w}$ be linearly independent vectors in $\R^n$. Prove that the vectors $A\mathbf{v}$ and $A\mathbf{w}$ are linearly independent.

Proof.

Suppose that we have a linear combination
\[c_1(A\mathbf{v})+c_2(A\mathbf{w})=\mathbf{0},\] where $c_1, c_2$ are scalars.
Out goal is to show that $c_1=c_2=0$.
Factoring out $A$, we have
\[A(c_1\mathbf{v}+c_2\mathbf{w})=\mathbf{0}.\]

Note that since $A$ is a nonsingular matrix, the equation $A\mathbf{x}=\mathbf{0}$ has only the zero solution $\mathbf{x}=\mathbf{0}$.
The above equality yields that $c_1\mathbf{v}+c_2\mathbf{w}$ is a solution to $A\mathbf{x}=\mathbf{0}$.
Hence, we have
\[c_1\mathbf{v}+c_2\mathbf{w}=\mathbf{0}.\]

By assumption, the vectors $\mathbf{v}$ and $\mathbf{w}$ are linearly independent, and this implies that $c_1=c_2=0$.

We have shown that whenever we have $c_1(A\mathbf{v})+c_2(A\mathbf{w})=\mathbf{0}$, we must have $c_1=c_2=0$. This yields that the vectors $A\mathbf{v}$ and $A\mathbf{w}$ are linearly independent.

Common Mistake

This is a midterm exam problem of Lienar Algebra at the Ohio State University.

One common mistake is to ignore the logic and write down whatever you know.
For example, some students started with $c_1\mathbf{v}+c_2\mathbf{w}=\mathbf{0}$ and since $\mathbf{v}$ and $\mathbf{w}$ are linearly independent, we have $c_1=c_2=0$.
Multiplying by $A$, we have $c_1A\mathbf{v}+c_2A\mathbf{w}=\mathbf{0}$ and $c_1=c_2=0$.

This argument is totally wrong.

The above argument is wrong because it started with different vectors than we want to prove to be linearly independent.
There is nothing wrong about the mathematical operations in the above arguments. However, that argument does not prove that the vectors $A\mathbf{v}$ and $A\mathbf{w}$ are linearly independent.

We should first assume that $c_1A\mathbf{v}+c_2A\mathbf{w}=\mathbf{0}$ and prove that $c_1=c_2=0$.

(a) Find a $3\times 3$ nonsingular matrix $A$ satisfying $3A=A^2+AB$, where \[B=\begin{bmatrix}
2 & 0 & -1 \\
0 &2 &-1 \\
-1 & 0 & 1
\end{bmatrix}.\]

(b) Find the inverse matrix of $A$.

Solution

(a) Find a $3\times 3$ nonsingular matrix $A$.

Assume that $A$ is nonsingular. Then the inverse matrix $A^{-1}$ exists.
Multiplying the given equality by $A^{-1}$ on the left, we obtain
\[3A^{-1}A=A^{-1}(A^2+AB)=A^{-1}A^2+A^{-1}AB=A+IB=A+B.\] Note that the left most term is equal to $3I$.
Hence, we have $3I=A+B$. Solving for $A$, we have
\[A=3I-B=\begin{bmatrix}
3 & 0 & 0 \\
0 &3 &0 \\
0 & 0 & 3
\end{bmatrix}-\begin{bmatrix}
2 & 0 & -1 \\
0 &2 &-1 \\
-1 & 0 & 1
\end{bmatrix}=\begin{bmatrix}
1 & 0 & 1 \\
0 &1 &1 \\
1 & 0 & 2
\end{bmatrix}.\]

As we see in part (b), this matrix is actually invertible.

(b) Find the inverse matrix of $A$.

To find the inverse matrix of $A$, we reduce the augmented matrix $[A\mid I]$ as follows:
\begin{align*}
[A\mid I]= \left[\begin{array}{rrr|rrr}
1 & 0 & 1 & 1 &0 & 0 \\
0 & 1 & 1 & 0 & 1 & 0 \\
1 & 0 & 2 & 0 & 0 & 1 \\
\end{array} \right] \xrightarrow{R_3-R_2}
\left[\begin{array}{rrr|rrr}
1 & 0 & 1 & 1 &0 & 0 \\
0 & 1 & 1 & 0 & 1 & 0 \\
0 & 0 & 1 & -1 & 0 & 1 \\
\end{array} \right]\\[6pt] \xrightarrow[R_2-R_3]{R_1-R_3}
\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & 2 &0 & -1 \\
0 & 1 & 0 & 1 & 1 & -1 \\
0 & 0 & 1 & -1 & 0 & 1 \\
\end{array} \right].
\end{align*}
The left 3 by 3 part is now the identity matrix.
So the inverse matrix is given by the right 3 by 3 part:
\[A^{-1}=\begin{bmatrix}
2 & 0 & -1 \\
1 &1 &-1 \\
-1 & 0 & 1
\end{bmatrix}.\]

Common Mistake

This is a midterm exam problem of Lienar Algebra at the Ohio State University.

One common mistake is: $A=3-B$. Don’t forget the identity matrix $I$. $3$ is a number but $3I$ is a matrix.
Also $3I$ is the 3 by 3 matrix whose diagonal entries are 3 and 0 elsewhere. Note that $3I$ is not the 3 by3 matrix whose entries are all 3.

Let $A$ and $B$ be $3\times 3$ matrices and let $C=A-2B$.
If
\[A\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}=B\begin{bmatrix}
2 \\
6 \\
10
\end{bmatrix},\] then is the matrix $C$ nonsingular? If so, prove it. Otherwise, explain why not.

Solution.

Using the given equality, we have
\begin{align*}
\mathbf{0}&=A\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}-B\begin{bmatrix}
2 \\
6 \\
10
\end{bmatrix}\\[6pt] &=A\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}-2B\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}\\[6pt] &=(A-2B)\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}\\[6pt] &=C\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}.
\end{align*}
Thus, we obtain
\[C\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}=\mathbf{0}.\] This implies that the vector $\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}$ is a nonzero solution to the equation $C\mathbf{x}=\mathbf{0}$.
Therefore, the matrix $C$ is singular.

Common Mistake

This is a midterm exam problem of Lienar Algebra at the Ohio State University.

One common mistake is that once you have
\[A\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}=2B\begin{bmatrix}
1 \\
3\\
5
\end{bmatrix},\] then you conclude that $A=2B$, and hence $C$ is the zero matrix.

This is WRONG!! You cannot cancel matrices, like numbers!

Another common mistake is to state something like “Because $C$ has a nonzero solution…”
There is no “solution” to a matrix. We can talk about the solution of a system, for example.

Determine whether the following matrices are nonsingular or not.

(a) $A=\begin{bmatrix}
1 & 0 & 1 \\
2 &1 &2 \\
1 & 0 & -1
\end{bmatrix}$.

(b) $B=\begin{bmatrix}
2 & 1 & 2 \\
1 &0 &1 \\
4 & 1 & 4
\end{bmatrix}$.

Solution.

Recall that an $n\times n$ matrix is nonsingular if $A\mathbf{x}=\mathbf{0}$ has only the zero solution. This is equivalent to the condition that the rank of $A$ is $n$.

(a) Is $A=\begin{bmatrix}
1 & 0 & 1 \\
2 &1 &2 \\
1 & 0 & -1
\end{bmatrix}$ nonsingular?

We find the rank of the matrix $A$ as follows.
\begin{align*}
A=\begin{bmatrix}
1 & 0 & 1 \\
2 &1 &2 \\
1 & 0 & -1
\end{bmatrix}
\xrightarrow[R_3-R_1]{R_2-2R_1} \begin{bmatrix}
1 & 0 & 1 \\
0 &1 &0 \\
0 & 0 & -2
\end{bmatrix}
\xrightarrow{-\frac{1}{2}R_3}
\begin{bmatrix}
1 & 0 & 1 \\
0 &1 &0 \\
0 & 0 & 1
\end{bmatrix}
\xrightarrow{R_1-R_3}
\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & 1
\end{bmatrix}.
\end{align*}
The last matrix is in reduced row echelon form and has three nonzero rows.
Hence, the rank of $A$ is $3$.
It follows that the matrix $A$ is nonsingular.

(b) Is $B=\begin{bmatrix}
2 & 1 & 2 \\
1 &0 &1 \\
4 & 1 & 4
\end{bmatrix}$ nonsingular?

Next, we compute the rank of $B$ as follows.
\begin{align*}
B=\begin{bmatrix}
2 & 1 & 2 \\
1 &0 &1 \\
4 & 1 & 4
\end{bmatrix}
\xrightarrow{R_1 \leftrightarrow R_2}
\begin{bmatrix}
1 & 0 & 1 \\
2 &1 &2 \\
4 & 1 & 4
\end{bmatrix}
\xrightarrow[R_3-4R_1]{R_2-2R_1}
\begin{bmatrix}
1 & 0 & 1 \\
0 &1 &0 \\
0 & 1 & 0
\end{bmatrix}
\xrightarrow{R_3-R_2}
\begin{bmatrix}
1 & 0 & 1 \\
0 &1 &0 \\
0 & 0 & 0
\end{bmatrix}.
\end{align*}

This computation shows that the rank of $B$ is $2$.
Hence, the matrix $B$ is singular.

Determine the values of a real number $a$ such that the matrix
\[A=\begin{bmatrix}
3 & 0 & a \\
2 &3 &0 \\
0 & 18a & a+1
\end{bmatrix}\] is nonsingular.
 

Solution.

We apply elementary row operations and obtain:
\begin{align*}
A=\begin{bmatrix}
3 & 0 & a \\
2 &3 &0 \\
0 & 18a & a+1
\end{bmatrix}
\xrightarrow{R_1-R_2}
\begin{bmatrix}
1 & -3 & a \\
2 &3 &0 \\
0 & 18a & a+1
\end{bmatrix}\\[6pt] \xrightarrow{R_2-2R_1}
\begin{bmatrix}
1 & -3 & a \\
0 &9 &-2a \\
0 & 18a & a+1
\end{bmatrix}
\xrightarrow{R_3-(2a)R_2}
\begin{bmatrix}
1 & -3 & a \\
0 &9 &-2a \\
0 & 0 & 4a^2+a+1
\end{bmatrix}.
\end{align*}

From this, we see that the matrix $A$ is nonsingular if and only if the $(3, 3)$-entry $4a^2+a+1$ is not zero.
By the quadratic formula, we see that
\[a=\frac{-1\pm \sqrt{-15}}{8}\] are solutions of $4a^2+a+1=0$.

Note that these are not real numbers. Thus, for any real number $a$, we have $4a^2+a+1\neq 0$.

Hence, we can divide the third row by this number, and eventually we can reduce it to the identity matrix.
So the rank of $A$ is $3$, and $A$ is nonsingular for any real number $a$.

(a) Suppose that a $3\times 3$ system of linear equations is inconsistent. Is the coefficient matrix of the system nonsingular?

(b) Suppose that a $3\times 3$ homogeneous system of linear equations has a solution $x_1=0, x_2=-3, x_3=5$. Is the coefficient matrix of the system nonsingular?

(c) Let $A$ be a $4\times 4$ matrix and let
\[\mathbf{v}=\begin{bmatrix}
1 \\
2 \\
3 \\
4
\end{bmatrix} \text{ and } \mathbf{w}=\begin{bmatrix}
4 \\
3 \\
2 \\
1
\end{bmatrix}.\] Suppose that we have $A\mathbf{v}=A\mathbf{w}$. Is the matrix $A$ nonsingular?

Hint.

Recall the following fact.

Let $A$ be an $n\times n$ matrix.

Then $A\mathbf{x}=\mathbf{b}$ has a unique solution for every $n\times 1$ column vector $\mathbf{b}$ if and only if $A$ is nonsingular.

Solution.

(a) Suppose that a $3\times 3$ system of linear equations is inconsistent. Is the coefficient matrix of the system nonsingular?

Let $A\mathbf{x}=\mathbf{b}$ be the system, where $A$ is the coefficient matrix and $\mathbf{b}$ is the constant term vector. Because this system is inconsistent, then $A$ is singular. (If $A$ would be nonsingular, then the system has a unique solution.)

(b) Suppose that a $3\times 3$ homogeneous system of linear equations has a solution $x_1=0, x_2=-3, x_3=5$. Is the coefficient matrix of the system nonsingular?

Recall that a homogeneous system of linear equations has always the zero solution. As the system has another solution $x_1=0, x_2=-3, x_3=5$, the system $A\mathbf{x}=\mathbf{0}$ must have infinitely many solutions. Here $A$ is the coefficient matrix.
Thus, the coefficient matrix $A$ is singular. (If $A$ would be nonsingular, the system has only one solution, which must be the zero solution.)

(c) Suppose that we have $A\mathbf{v}=A\mathbf{w}$. Is the matrix $A$ nonsingular?

Because $A\mathbf{v}=A\mathbf{w}$, we have
\[A(\mathbf{v}-\mathbf{w})=A\mathbf{v}-A\mathbf{w}=\mathbf{0}.\] Note that
\[\mathbf{v}-\mathbf{w}=\begin{bmatrix}
1 \\
2 \\
3 \\
4
\end{bmatrix}-\begin{bmatrix}
4 \\
3 \\
2 \\
1
\end{bmatrix}=\begin{bmatrix}
-3 \\
-1 \\
1 \\
3
\end{bmatrix}\] is a nonzero vector.
This implies that the homogeneous system $A\mathbf{x}=\mathbf{0}$ has infinitely many solutions as we have found a nonzero solution.
Hence, $A$ is singular.

Suppose that $M, P$ are two $n \times n$ non-singular matrix. Prove that there is a matrix $N$ such that $MN = P$.

Proof.

As non-singularity and invertibility are equivalent, we know that $M$ has the inverse matrix $M^{-1}$.

Let us think backwards. Suppose that we have $MN=P$ for some matrix $N$, which we want to find.
Then multiply $M^{-1}$ on the left of the equation $MN = P$ yields $N = M^{-1} P$.

This is the matrix we are looking for, as $ M N = M ( M^{-1} P) = P$.

Suppose that $B=\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis for $\R^2$. Let $S:=[\mathbf{v}_1, \mathbf{v}_2]$.
Note that as the column vectors of $S$ are linearly independent, the matrix $S$ is invertible.

Prove that for each vector $\mathbf{v} \in V$, the vector $S^{-1}\mathbf{v}$ is the coordinate vector of $\mathbf{v}$ with respect to the basis $B$.

Proof.

We first express the vector $\mathbf{v}$ as a linear combination of the basis vectors
\[\mathbf{v}=c_1\mathbf{v}_+c_2 \mathbf{v}_2.\] This expression is unique and the coordinate vector of $\mathbf{v}$ with respect to the basis $B$ is defined to be
\[[\mathbf{v}]_B =\begin{bmatrix}
c_1 \\
c_2
\end{bmatrix}.\]

Let
\[S^{-1}\mathbf{v}= \begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}.\] Or equivalently,
\[\mathbf{v}=S\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}.\] Our goal is to show that $\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}
=
\begin{bmatrix}
c_1 \\
c_2
\end{bmatrix}$.

We have
\begin{align*}
c_1\mathbf{v}_+c_2 \mathbf{v}_2&=\mathbf{v}=S\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}=x_1\mathbf{v}_1+x_2\mathbf{v}_2.
\end{align*}

Hence
\[(x_1-c_1)\mathbf{v}_1+(x_2-c_2)\mathbf{v}_2=\mathbf{0}.\] As $B=\{\mathbf{v}_1, \mathbf{v}_2\}$ is linearly independent, we obtain $x_1=c_1$ and $x_2=c_2$.

Therefore,
\[S^{-1}\mathbf{v}=\begin{bmatrix}
c_1 \\
c_2
\end{bmatrix}=[\mathbf{v}]_B.\] Click here if solved 8The post The Vector $S^{-1}\mathbf{v}$ is the Coordinate Vector of $\mathbf{v}$ first appeared on Problems in Mathematics.]]> https://yutsumura.com/the-vector-s-1mathbfv-is-the-coordinate-vector-of-mathbfv/feed/ 0 6253 https://yutsumura.com/diagonalize-a-2-by-2-symmetric-matrix/ https://yutsumura.com/diagonalize-a-2-by-2-symmetric-matrix/#comments Sun, 17 Dec 2017 02:58:29 +0000 https://yutsumura.com/?p=6226 Diagonalize the $2\times 2$ matrix $A=\begin{bmatrix} 2 & -1\\ -1& 2 \end{bmatrix}$ by finding a nonsingular matrix $S$ and a diagonal matrix $D$ such that $S^{-1}AS=D$.   Solution. The characteristic polynomial $p(t)$ of the...