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.

The post Determine whether the Given 3 by 3 Matrices are Nonsingular first appeared on Problems in Mathematics.

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 $a$.

The post For What Values of $a$, Is the Matrix Nonsingular? first appeared on Problems in Mathematics.

If $A, B$ have the same rank, can we conclude that they are row-equivalent?

If so, then prove it. If not, then provide a counterexample.

Solution.

Having the same rank does not mean they are row-equivalent.

For a simple counterexample, consider $A = \begin{bmatrix} 1 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 1 \end{bmatrix}$.

Both of these matrices have rank 1, but are not row-equivalent because they are already in reduced row echelon form.

Another solution.

The problem doesn’t specify the sizes of matrices $A$, $B$.

Note that if the sizes of $A$ and $B$ are distinct, then they can never be row-equivalent.
Keeping this in mind, let us consider the following two matrices.

\[A=\begin{bmatrix}
1 \\
0
\end{bmatrix} \text{ and } B=\begin{bmatrix}
1 & 0
\end{bmatrix}.\] Then both matrices are in reduced row echelon form and have rank $1$.
As noted above, they are not row-equivalent because the sizes are distinct.

The post If Two Matrices Have the Same Rank, Are They Row-Equivalent? first appeared on Problems in Mathematics.

For each of the following matrices, find a row-equivalent matrix which is in reduced row echelon form. Then determine the rank of each matrix.

(a) $A = \begin{bmatrix} 1 & 3 \\ -2 & 2 \end{bmatrix}$.

(b) $B = \begin{bmatrix} 2 & 6 & -2 \\ 3 & -2 & 8 \end{bmatrix}$.

(c) $C = \begin{bmatrix} 2 & -2 & 4 \\ 4 & 1 & -2 \\ 6 & -1 & 2 \end{bmatrix}$.

(d) $D = \begin{bmatrix} -2 \\ 3 \\ 1 \end{bmatrix}$.

(e) $E = \begin{bmatrix} -2 & 3 & 1 \end{bmatrix}$.

Definition (Rank of a Matrix).

The rank of a matrix $A$ is the number of nonzero rows in the reduced row echelon form matrix $\rref(A)$ that is row equivalent to $A$.

Solution.

(a) $A = \begin{bmatrix} 1 & 3 \\ -2 & 2 \end{bmatrix}$

The matrix $A$ has rank 2, which can be seen by computing
\[ \begin{bmatrix} 1 & 3 \\ -2 & 2 \end{bmatrix} \xrightarrow{R_2 + 2 R_1} \begin{bmatrix} 1 & 3 \\ 0 & 8 \end{bmatrix} \xrightarrow{\frac{1}{8} R_2 } \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} \xrightarrow{R_1 – 3 R_2} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. \] Because the row-reduced matrix has two non-zero rows, the rank of $A$ is $2$.

(b) $B = \begin{bmatrix} 2 & 6 & -2 \\ 3 & -2 & 8 \end{bmatrix}$

The matrix $B$ has rank 2, which can be seen by computing
\begin{align*}
\begin{bmatrix} 2 & 6 & -2 \\ 3 & -2 & 8 \end{bmatrix} \xrightarrow{\frac{1}{2} R_1 } \begin{bmatrix} 1 & 3 & -1 \\ 3 & -2 & 8 \end{bmatrix} \xrightarrow{R_2 – 3 R_1} \begin{bmatrix} 1 & 3 & -1 \\ 0 & -11 & 11 \end{bmatrix} \\[6pt] \xrightarrow{\frac{-1}{11} R_2 } \begin{bmatrix} 1 & 3 & -1 \\ 0 & 1 & -1 \end{bmatrix} \xrightarrow{R_1 – 3 R_2} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & -1 \end{bmatrix}
\end{align*}

(c) $C = \begin{bmatrix} 2 & -2 & 4 \\ 4 & 1 & -2 \\ 6 & -1 & 2 \end{bmatrix}$

The matrix $C$ has rank 2, which can be seen by computing
\begin{align*}
\begin{bmatrix} 2 & -2 & 4 \\ 4 & 1 & -2 \\ 6 & -1 & 2 \end{bmatrix} \xrightarrow{\frac{1}{2} R_1 } \begin{bmatrix} 1 & -1 & 2 \\ 4 & 1 & -2 \\ 6 & -1 & 2 \end{bmatrix} \xrightarrow[R_3 – 6 R_1]{R_2 – 4 R_1} \begin{bmatrix} 1 & -1 & 2 \\ 0 & 5 & -10 \\ 0 & 5 & -10 \end{bmatrix} \\[6pt] \xrightarrow{R_3 – R_2} \begin{bmatrix} 1 & -1 & 2 \\ 0 & 5 & -10 \\ 0 & 0 & 0 \end{bmatrix} \xrightarrow{ \frac{1}{5} R_2 } \begin{bmatrix} 1 & -1 & 2 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{bmatrix} \xrightarrow{R_1 + R_2} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{bmatrix} .
\end{align*}

(d) $D = \begin{bmatrix} -2 \\ 3 \\ 1 \end{bmatrix}$

The matrix $D$ has rank 1, which can be seen by calculating:
\[\begin{bmatrix} -2 \\ 3 \\ 1 \end{bmatrix} \xrightarrow{ \frac{-1}{2} R_1 } \begin{bmatrix} 1 \\ 3 \\1 \end{bmatrix} \xrightarrow{R_2 – 3 R_1 , R_3 – R_1} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}.\]

(e) $E = \begin{bmatrix} -2 & 3 & 1 \end{bmatrix}$

The matrix $E$ has rank 1, which can be seen by calculating:
\[\begin{bmatrix} -2 & 3 & 1 \end{bmatrix} \xrightarrow{ \frac{-1}{2} R_1 } \begin{bmatrix} 1 & \frac{-3}{2} & \frac{-1}{2} \end{bmatrix}.\] Click here if solved 134

The post Find a Row-Equivalent Matrix which is in Reduced Row Echelon Form and Determine the Rank first appeared on Problems in Mathematics.

If $A, B, C$ are three $m \times n$ matrices such that $A$ is row-equivalent to $B$ and $B$ is row-equivalent to $C$, then can we conclude that $A$ is row-equivalent to $C$?

If so, then prove it. If not, then provide a counterexample.

Definition (Row Equivalent).

Two matrices are said to be row equivalent if one can be obtained from the other by a sequence of elementary row operations.

Proof.

Yes, in this case $A$ and $C$ are row-equivalent.

By assumption, the matrices $A$ and $B$ are row-equivalent, which means that there is a sequence of elementary row operations that turns $A$ into $B$.

Call this sequence $r_1 , r_2 , \cdots , r_n$, where each $r_i$ is an elementary row operation.
(Start with applying $r_1$ to $A$.)

By another assumption, $B$ is row-equivalent to $C$, which means that there is a sequence of elementary row operations which transforms $B$ into $C$; call this sequence $s_1 , s_2 , \cdots , s_m$.

Putting these sequences together, the operations $r_1 , r_2 , \cdots , r_n$ , $s_1 , s_2 , \cdots , s_m$ will transform the matrix $A$ into $C$.

This proves that $A$ and $C$ are row-equivalent.

The post Row Equivalence of Matrices is Transitive first appeared on Problems in Mathematics.

For an $m\times n$ matrix $A$, we denote by $\mathrm{rref}(A)$ the matrix in reduced row echelon form that is row equivalent to $A$.
For example, consider the matrix $A=\begin{bmatrix}
1 & 1 & 1 \\
0 &2 &2
\end{bmatrix}$
Then we have
\[A=\begin{bmatrix}
1 & 1 & 1 \\
0 &2 &2
\end{bmatrix}
\xrightarrow{\frac{1}{2}R_2}
\begin{bmatrix}
1 & 1 & 1 \\
0 &1 & 1
\end{bmatrix}
\xrightarrow{R_1-R_2}
\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &1
\end{bmatrix}\] and the last matrix is in reduced row echelon form.
Hence $\mathrm{rref}(A)=\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &1
\end{bmatrix}$.

Find an example of matrices $A$ and $B$ such that
\[\mathrm{rref}(AB)\neq \mathrm{rref}(A) \mathrm{rref}(B).\]

Proof.

Let
\[A=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix} \text{ and } B=\begin{bmatrix}
0 & 0\\
1& 0
\end{bmatrix}.\] Then $A$ is already in reduced row echelon from.
So we have $\mathrm{rref}(A)=A$.

Applying the elementary row operations, we obtain
\[B=\begin{bmatrix}
0 & 0\\
1& 0
\end{bmatrix} \xrightarrow{R_1\leftrightarrow R_2}\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix}.\] As the last matrix is in reduced row echelon from, we have $\mathrm{rref}(B)=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix}$.
Therefore, we see that
\[\mathrm{rref}(A) \mathrm{rref}(B)=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix}=\begin{bmatrix}
0 & 0\\
0& 0
\end{bmatrix}.\]

The product of $A$ and $B$ is
\[AB=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}
\begin{bmatrix}
0 & 0\\
1& 0
\end{bmatrix}=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix}.\] It follows that $\mathrm{rref}(AB)=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix}$.

In summary, we have
\[\mathrm{rref}(AB)=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix} \neq
\begin{bmatrix}
0 & 0\\
0& 0
\end{bmatrix}
=\mathrm{rref}(A) \mathrm{rref}(B),\] as required.

The post An Example of Matrices $A$, $B$ such that $\mathrm{rref}(AB)\neq \mathrm{rref}(A) \mathrm{rref}(B)$ first appeared on Problems in Mathematics.

For each of the following $3\times 3$ matrices $A$, determine whether $A$ is invertible and find the inverse $A^{-1}$ if exists by computing the augmented matrix $[A|I]$, where $I$ is the $3\times 3$ identity matrix.

(a) $A=\begin{bmatrix}
1 & 3 & -2 \\
2 &3 &0 \\
0 & 1 & -1
\end{bmatrix}$
 
(b) $A=\begin{bmatrix}
1 & 0 & 2 \\
-1 &-3 &2 \\
3 & 6 & -2
\end{bmatrix}$.

Elementary Row Operations and Inverse Matrices

Recall the following procedure of testing the invertibility of $A$ as well as finding the inverse matrix if exists.
If the augmented matrix $[A|I]$ is transformed into a matrix of the form $[I|B]$, then the matrix $A$ is invertible and the inverse matrix $A^{-1}$ is given by $B$.
If the reduced row echelon form matrix for $[A|I]$ is not of the form $[I|B]$, then the matrix $A$ is not invertible.

Solution.

(a) $A=\begin{bmatrix}
1 & 3 & -2 \\
2 &3 &0 \\
0 & 1 & -1
\end{bmatrix}$

We apply the elementary row operations as follows.
We have
\begin{align*}
&[A|I]= \left[\begin{array}{rrr|rrr}
1 & 3 & -2 & 1 &0 & 0 \\
2 & 3 & 0 & 0 & 1 & 0 \\
0 & 1 & -1 & 0 & 0 & 1 \\
\end{array} \right] \xrightarrow{R_2-2R_1}
\left[\begin{array}{rrr|rrr}
1 & 3 & -2 & 1 &0 & 0 \\
0 & -3 & 4 & -2 & 1 & 0 \\
0 & 1 & -1 & 0 & 0 & 1 \\
\end{array} \right]\\[6pt] &\xrightarrow{R_2\leftrightarrow R_3}
\left[\begin{array}{rrr|rrr}
1 & 3 & -2 & 1 &0 & 0 \\
0 & 1 & -1 & 0 & 0 & 1 \\
0 & -3 & 4 & -2 & 1 & 0 \\
\end{array} \right] \xrightarrow{\substack{R_1-3R_2\\ R_3+3R_2}}
\left[\begin{array}{rrr|rrr}
1 & 0& 1 & 1 &0 & -3 \\
0 & 1 & -1 & 0 & 0 & 1 \\
0 & 0 & 1 & -2 & 1 & 3 \\
\end{array} \right]\\[6pt] &\xrightarrow{\substack{R_1-R_3\\ R_2+R_3}}
\left[\begin{array}{rrr|rrr}
1 & 0& 0 & 3 & -1 & -6 \\
0 & 1 & 0 & -2 & 1 & 4 \\
0 & 0 & 1 & -2 & 1 & 3 \\
\end{array} \right].
\end{align*}

The left $3\times 3$ part of the last matrix is the identity matrix.
This implies that $A$ is invertible and the inverse matrix is given by the right $3\times 3$ matrix.
Hence
\[A^{-1}=\begin{bmatrix}
3 & -1 & -6 \\
-2 &1 &4 \\
-2 & 1 & 3
\end{bmatrix}.\]

(b) $A=\begin{bmatrix}
1 & 0 & 2 \\
-1 &-3 &2 \\
3 & 6 & -2
\end{bmatrix}$.

Now we consider the matrix $A=\begin{bmatrix}
1 & 0 & 2 \\
-1 &-3 &2 \\
3 & 6 & -2
\end{bmatrix}$.

Applying elementary row operations, we obtain
\begin{align*}
&[A|I]= \left[\begin{array}{rrr|rrr}
1 & 0 & 2 & 1 &0 & 0 \\
-1 & -3 & 2 & 0 & 1 & 0 \\
3 & 6 & -2 & 0 & 0 & 1 \\
\end{array} \right] \xrightarrow{\substack{R_2+R_1\\ R_3-3R_1}}
\left[\begin{array}{rrr|rrr}
1 & 0 & 2 & 1 &0 & 0 \\
0 & -3 & 4 & 1 & 1 & 0 \\
0 & 6 & -8 & -3 & 0 & 1 \\
\end{array} \right]\\[6pt] &\xrightarrow{R_3-2R_1}
\left[\begin{array}{rrr|rrr}
1 & 0 & 2 & 1 &0 & 0 \\
0 & -3 & 4 & 1 & 1 & 0 \\
0 & 0 & 0 & -5 & -2 & 1 \\
\end{array} \right] \xrightarrow{\frac{-1}{3}R_2}
\left[\begin{array}{rrr|rrr}
1 & 0 & 2 & 1 &0 & 0 \\
0 & 1 & -4/3 & -1/3 & -1/3 & 0 \\
0 & 0 & 0 & -5 & -2 & 1 \\
\end{array} \right].
\end{align*}

The last matrix is in reduced row echelon form but the left $3\times 3$ part is not the identity matrix $I$.
It follows that the matrix $A$ is not invertible.

Related Question.

This is a linear algebra exam problem at the Ohio State University.

The solution is given in the post↴
Find the Inverse Matrix of a $3\times 3$ Matrix if Exists

The post Find the Inverse Matrices if Matrices are Invertible by Elementary Row Operations first appeared on Problems in Mathematics.

Consider the following system of linear equations
\begin{align*}
2x+3y+z&=-1\\
3x+3y+z&=1\\
2x+4y+z&=-2.
\end{align*}

(a) Find the coefficient matrix $A$ for this system.

(b) Find the inverse matrix of the coefficient matrix found in (a)

(c) Solve the system using the inverse matrix $A^{-1}$.

Solution.

(a) Find the coefficient matrix $A$ for this system.

The system can be written as
\[A\mathbf{x}=\mathbf{b},\] where
\[A=\begin{bmatrix}
2 & 3 & 1 \\
3 &3 &1 \\
2 & 4 & 1
\end{bmatrix}\] is the coefficient matrix of the system and
\[\mathbf{x}=\begin{bmatrix}
x \\
y \\
z
\end{bmatrix} \text{ and } \mathbf{b}=\begin{bmatrix}
-1 \\
1 \\
-2
\end{bmatrix}.\]

(b) Find the inverse matrix of the coefficient matrix

We apply the elementary row operations to the augmented matrix $[A\mid I]$, where $I$ is the $3\times 3$ identity matrix.
\begin{align*}
&[A\mid I]= \left[\begin{array}{rrr|rrr}
2 & 3 & 1 & 1 &0 & 0 \\
3 & 3 & 1 & 0 & 1 & 0 \\
2 & 4 & 1 & 0 & 0 & 1 \\
\end{array} \right] \xrightarrow{\substack{R_2-R_1\\R_3-R_1}}
\left[\begin{array}{rrr|rrr}
2 & 3 & 1 & 1 &0 & 0 \\
1 & 0 & 0 & -1 & 1 & 0 \\
0 & 1 & 0 & -1 & 0 & 1 \\
\end{array} \right]\\[6pt] &\xrightarrow[\text{then } R_2 \leftrightarrow R_3]{R_1\leftrightarrow R_2}
\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & -1 & 1 & 0 \\
0 & 1 & 0 & -1 & 0 & 1 \\
2 & 3 & 1 & 1 &0 & 0
\end{array} \right] \xrightarrow{R_3-2R_1}
\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & -1 & 1 & 0 \\
0 & 1 & 0 & -1 & 0 & 1 \\
0 & 3 & 1 & 3 & -2 & 0
\end{array} \right]\\[6pt] &\xrightarrow{R_3-3R_2}
\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & -1 & 1 & 0 \\
0 & 1 & 0 & -1 & 0 & 1 \\
0 & 0 & 1 & 6 & -2 & -3
\end{array} \right].
\end{align*}

Now the left $3\times 3$ part has become the identity matrix.
So the matrix $A$ is invertible and the inverse is given by the right $3\times 3$ part.
Hence we obtain
\[A^{-1}=\begin{bmatrix}
-1 & 1 & 0 \\
-1 &0 &1 \\
6 & -2 & -3
\end{bmatrix}.\]

(c) Solve the system using the inverse matrix $A^{-1}$.

As noted in (a), the system can be written using matrices as
\[A\mathbf{x}=\mathbf{b}.\] Multiplying by the inverse $A^{-1}$ on the left, we have
\begin{align*}
A^{-1}A\mathbf{x}=A^{-1}\mathbf{b}\\
I\mathbf{x}=A^{-1}\mathbf{b}\\
\mathbf{x}=A^{-1}\mathbf{b}.
\end{align*}
Therefore the solution $\mathbf{x}$ of the system is given by
\begin{align*}
\mathbf{x}&=A^{-1}\mathbf{b}\\
&=\begin{bmatrix}
-1 & 1 & 0 \\
-1 &0 &1 \\
6 & -2 & -3
\end{bmatrix}\begin{bmatrix}
-1 \\
1 \\
-2
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
2 \\
-1 \\
-2
\end{bmatrix}.
\end{align*}
Thus, the solution of the system is
\[x=2, y=-1, z=-2.\] Click here if solved 69

The post Solve the System of Linear Equations Using the Inverse Matrix of the Coefficient Matrix first appeared on Problems in Mathematics.

Find all the eigenvalues and eigenvectors of the matrix
\[A=\begin{bmatrix}
10001 & 3 & 5 & 7 &9 & 11 \\
1 & 10003 & 5 & 7 & 9 & 11 \\
1 & 3 & 10005 & 7 & 9 & 11 \\
1 & 3 & 5 & 10007 & 9 & 11 \\
1 &3 & 5 & 7 & 10009 & 11 \\
1 &3 & 5 & 7 & 9 & 10011
\end{bmatrix}.\]

(MIT, Linear Algebra Homework Problem)
 

Solution.

Let
\[B=A-10000I,\] where $I$ is the $6 \times 6$ identity matrix. That is, we have
\[B=\begin{bmatrix}
1 & 3 & 5 & 7 &9 & 11 \\
1 & 3 & 5 & 7 &9 & 11 \\
1 & 3 & 5 & 7 &9 & 11 \\
1 & 3 & 5 & 7 &9 & 11 \\
1 & 3 & 5 & 7 &9 & 11 \\
1 & 3 & 5 & 7 &9 & 11 \\
\end{bmatrix}.\]

Since all the rows are the same, the matrix $B$ is singular and hence $\lambda=0$ is an eigenvalue of $B$.
Let us determine the geometric multiplicity of $\lambda=0$ (namely, the dimension of the null space of $B$).

We apply elementary row operations to $B$ and obtain
\begin{align*}
B\xrightarrow{\text{elementary row operations}}
\begin{bmatrix}
1 & 3 & 5 & 7 &9 & 11 \\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
\end{bmatrix}.
\end{align*}
Thus, if $B\mathbf{x}=\mathbf{0}$, then we have
\[x_1=-3x_2-5x_3-7x_4-9x_5-11x_6.\]

It follows from this that basis vectors of the eigenspace $E_0=\calN(B)$ are
\[\begin{bmatrix}
-3 \\
1 \\
0 \\
0 \\
0\\
0
\end{bmatrix}, \begin{bmatrix}
-5 \\
0 \\
1 \\
0 \\
0\\
0
\end{bmatrix}, \begin{bmatrix}
-7 \\
0 \\
0 \\
1 \\
0\\
0
\end{bmatrix}, \begin{bmatrix}
-9 \\
0 \\
0 \\
0 \\
1\\
0
\end{bmatrix}, \begin{bmatrix}
-11 \\
0 \\
0 \\
0 \\
0\\
1
\end{bmatrix},\] and hence the geometric multiplicity corresponding to $\lambda=0$ is $5$.

By inspection, we see that
\[B\mathbf{v}=36\mathbf{v},\] where
\[\mathbf{v}=\begin{bmatrix}
1 \\
1 \\
1 \\
1 \\
1\\
1
\end{bmatrix}.\] Thus, it yields that $\lambda=36$ is an eigenvalue of $B$ and $\mathbf{v}$ is a corresponding eigenvector.

Recall that the algebraic multiplicity of an eigenvalue is greater than or equal to the geometric multiplicity.

Also the sum of algebraic multiplicities of all eigenvalues of $B$ is equal to $6$ since $B$ is a $6\times 6$ matrix.

It follows from this observation that we determine that the algebraic multiplicity of $\lambda=0$ is $5$ and the algebraic and geometric multiplicities of $\lambda=36$ are both $1$.
Hence the vector $\mathbf{v}$ form a basis of the eigenspace $E_{36}$.

Now that we have determined eigenvalues and eigenvectors of $B$, we can deduce those of $A$ as follows.

In general, if $A=B+cI$, then the eigenvalues of $A$ are $\lambda+c$, where $\lambda$ are eigenvalues of $B$.
The eigenvectors for $A$ corresponding to $\lambda+c$ are exactly the eigenvectors for $B$ corresponding $\lambda$.
(See the post “Eigenvalues and algebraic/geometric multiplicities of matrix $A+cI$” for a proof.)

In the current problem, we have $A=B+10000I$, and thus $c=10000$.
Therefore, the eigenvalues of $A$ are $10000, 10036$.
Eigenvectors corresponding to $10000$ are
\[x_2\begin{bmatrix}
-3 \\
1 \\
0 \\
0 \\
0\\
0
\end{bmatrix}+x_3\begin{bmatrix}
-5 \\
0 \\
1 \\
0 \\
0\\
0
\end{bmatrix}+x_4\begin{bmatrix}
-7 \\
0 \\
0 \\
1 \\
0\\
0
\end{bmatrix}+x_5\begin{bmatrix}
-9 \\
0 \\
0 \\
0 \\
1\\
0
\end{bmatrix}+x_6\begin{bmatrix}
-11 \\
0 \\
0 \\
0 \\
0\\
1
\end{bmatrix},\] where $(x_2, x_3, x_4, x_5, x_6)\neq (0, 0, 0, 0, 0, 0)$.

The eigenvector corresponding to $10036$ is
\[a\begin{bmatrix}
1 \\
1 \\
1 \\
1 \\
1\\
1
\end{bmatrix},\] where $a$ is any nonzero scalar.

The post Find All the Eigenvalues and Eigenvectors of the 6 by 6 Matrix first appeared on Problems in Mathematics.

Let $T$ be the linear transformation from the $3$-dimensional vector space $\R^3$ to $\R^3$ itself satisfying the following relations.
\begin{align*}
T\left(\, \begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix} \,\right)
=\begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}, \qquad T\left(\, \begin{bmatrix}
2 \\
3 \\
5
\end{bmatrix} \, \right) =
\begin{bmatrix}
0 \\
2 \\
-1
\end{bmatrix}, \qquad
T \left( \, \begin{bmatrix}
0 \\
1 \\
2
\end{bmatrix} \, \right)=
\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}.
\end{align*}
Then for any vector
\[\mathbf{x}=\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}\in \R^3,\] find the formula for $T(\mathbf{x})$.

 

 
We give two solutions using two different methods.

Solution 1 using the matrix representation.

The first solution uses the matrix representation of $T$.
Let $A$ be the matrix representation of the linear transformation $T$ with respect to the standard basis of $\R^3$.
Then we have $T(\mathbf{x})=A\mathbf{x}$ by definition.
We determine the matrix $A$ as follows.
We have
\begin{align*}
A\begin{bmatrix}
1 & 2 & 0 \\
1 &3 &1 \\
1 & 5 & 2
\end{bmatrix}&= \begin{bmatrix}
A\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}, & A\begin{bmatrix}
2 \\
3 \\
5
\end{bmatrix}, & A\begin{bmatrix}
0 \\
1 \\
2
\end{bmatrix} \\
\end{bmatrix}\\[6 pt] &=\begin{bmatrix}
T\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}, & T\begin{bmatrix}
2 \\
3 \\
5
\end{bmatrix}, & T\begin{bmatrix}
0 \\
1 \\
2
\end{bmatrix} \\
\end{bmatrix}
=\begin{bmatrix}
1 & 0 & 1 \\
0 &2 &0 \\
1 & -1 & 0
\end{bmatrix}.
\end{align*}
It follows that we have
\begin{align*}
A=\begin{bmatrix}
1 & 0 & 1 \\
0 &2 &0 \\
1 & -1 & 0
\end{bmatrix}
\begin{bmatrix}
1 & 2 & 0 \\
1 &3 &1 \\
1 & 5 & 2
\end{bmatrix}^{-1}.
\end{align*}

Let us find the inverse matrix by using augmented matrix.
\begin{align*}
\left[\begin{array}{rrr|rrr}
1 & 2 & 0 & 1 &0 & 0 \\
1 & 3 & 1 & 0 & 1 & 0 \\
1 & 5 & 2 & 0 & 0 & 1 \\
\end{array} \right] \xrightarrow{\substack{R_2-R_1\\R_3-R_1}}
\left[\begin{array}{rrr|rrr}
1 & 2 & 0 & 1 &0 & 0 \\
0 & 1 & 1 & -1 & 1 & 0 \\
0 & 3 & 2 & -1 & 0 & 1 \\
\end{array} \right]\\[6 pt] \xrightarrow{\substack{R_1-2R_2 \\R_3-3R_2}}
\left[\begin{array}{rrr|rrr}
1 & 0 & -2 & 3 &-2 & 0 \\
0 & 1 & 1 & -1 & 1 & 0 \\
0 & 0 & -1 & 2 & -3 & 1 \\
\end{array} \right] \xrightarrow{-R_3}
\left[\begin{array}{rrr|rrr}
1 & 0 & -2 & 3 &-2 & 0 \\
0 & 1 & 1 & -1 & 1 & 0 \\
0 & 0 & 1 & -2 & 3 & -1 \\
\end{array} \right]\\[6 pt] \xrightarrow{\substack{R_1+2R_3\\ R_2-R_3}}
\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & -1 &4 & -2 \\
0 & 1 & 0 & 1 & -2 & 1 \\
0 & 0 & 1 & -2 & 3 & -1 \\
\end{array} \right].
\end{align*}
Thus we obtain the inverse matrix
\[\begin{bmatrix}
1 & 2 & 0 \\
1 &3 &1 \\
1 & 5 & 2
\end{bmatrix}^{-1}=\begin{bmatrix}
-1 & 4 & -2 \\
1 &-2 &1 \\
-2 & 3 & -1
\end{bmatrix},\] and hence we have
\begin{align*}
A=\begin{bmatrix}
1 & 0 & 1 \\
0 &2 &0 \\
1 & -1 & 0
\end{bmatrix}
\begin{bmatrix}
-1 & 4 & -2 \\
1 &-2 &1 \\
-2 & 3 & -1
\end{bmatrix}
=\begin{bmatrix}
-3 & 7 & -3 \\
2 &-4 &2 \\
-2 & 6 & -3
\end{bmatrix}.
\end{align*}
Using the relation $T(\mathbf{x})=A\mathbf{x}$, the formula for $T(\mathbf{x})$ is obtained as follows.
\begin{align*}
T(\mathbf{x})&=A\mathbf{x}\\
&=\begin{bmatrix}
-3 & 7 & -3 \\
2 &-4 &2 \\
-2 & 6 & -3
\end{bmatrix}\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}\\[6 pt] &=\begin{bmatrix}
-3x+7y-3z \\
2x-4y+2z \\
-2x+6y-3z
\end{bmatrix}.
\end{align*}

Solution 2 using a linear combination and linearity.

The second method is to find the linear combination
\[\mathbf{x}=c_1\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}+c_2\begin{bmatrix}
2 \\
3 \\
5
\end{bmatrix}+c_3\begin{bmatrix}
0 \\
1 \\
2
\end{bmatrix}\] and use the linearity of the linear transformation.
To find the coefficients $c_1, c_2, c_3$, we consider the augmented matrix
\[ \left[\begin{array}{rrr|r}
1 & 2 & 0 & x \\
1 &3 & 1 & y \\
1 & 5 & 2 & z
\end{array} \right]\] and we reduce this matrix by elementary row operations.
The reduction operations are exactly the same as in solution 1 and we obtain
\begin{align*}
c_1&=-x+4y-2z\\
c_2&=x-2y+z\\
c_3&=-2x+3y-z.
\end{align*}

Therefore, we have by the linearity of $T$
\begin{align*}
T(\mathbf{x})&=T\left(\,c_1\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}+c_2\begin{bmatrix}
2 \\
3 \\
5
\end{bmatrix}+c_3\begin{bmatrix}
0 \\
1 \\
2
\end{bmatrix} \, \right)\\[6 pt] &=c_1T\left(\,\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}\, \right)+c_2T\left(\,\begin{bmatrix}
2 \\
3 \\
5
\end{bmatrix}\, \right)+c_3T\left(\,\begin{bmatrix}
0 \\
1 \\
2
\end{bmatrix} \, \right)\\[6 pt] &=(-x+4y-2z)\begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}+(x-2y+z)\begin{bmatrix}
0 \\
2 \\
-1
\end{bmatrix} +(-2x+3y-z)\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}\\[6 pt] &=\begin{bmatrix}
-3x+7y-3z \\
2x-4y+2z \\
-2x+6y-3z
\end{bmatrix}.
\end{align*}

Related Question.

A similar problem with a linear transformation from $\R^2$ to $\R^3$ is given in the post
“Give a formula for a linear transformation from $\R^2$ to $\R^3$“.

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