# Determine Whether Given Matrices are Similar

## Problem 391

**(a)** Is the matrix $A=\begin{bmatrix}

1 & 2\\

0& 3

\end{bmatrix}$ similar to the matrix $B=\begin{bmatrix}

3 & 0\\

1& 2

\end{bmatrix}$?

**(b)** Is the matrix $A=\begin{bmatrix}

0 & 1\\

5& 3

\end{bmatrix}$ similar to the matrix $B=\begin{bmatrix}

1 & 2\\

4& 3

\end{bmatrix}$?

**(c)** Is the matrix $A=\begin{bmatrix}

-1 & 6\\

-2& 6

\end{bmatrix}$ similar to the matrix $B=\begin{bmatrix}

3 & 0\\

0& 2

\end{bmatrix}$?

**(d)** Is the matrix $A=\begin{bmatrix}

-1 & 6\\

-2& 6

\end{bmatrix}$ similar to the matrix $B=\begin{bmatrix}

1 & 2\\

-1& 4

\end{bmatrix}$?

Contents

- Problem 391
- Solution.
- (a) Is the matrix $A=\begin{bmatrix} 1 & 2\\ 0& 3 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 3 & 0\\ 1& 2 \end{bmatrix}$?
- (b) Is the matrix $A=\begin{bmatrix} 0 & 1\\ 5& 3 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 1 & 2\\ 4& 3 \end{bmatrix}$?
- (c) Is the matrix $A=\begin{bmatrix} -1 & 6\\ -2& 6 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 3 & 0\\ 0& 2 \end{bmatrix}$?
- (d) Is the matrix $A=\begin{bmatrix} -1 & 6\\ -2& 6 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 1 & 2\\ -1& 4 \end{bmatrix}$?

- Related Question.

## Solution.

### (a) Is the matrix $A=\begin{bmatrix}

1 & 2\\

0& 3

\end{bmatrix}$ similar to the matrix $B=\begin{bmatrix}

3 & 0\\

1& 2

\end{bmatrix}$?

Recall that if $A$ and $B$ are similar, then their determinants are the same.

We compute

\begin{align*}

\det(A)=(1)(3)-(2)(0)=3 \text{ and } \det(B)=(3)(2)-(0)(1)=6.

\end{align*}

Thus, $\det(A)\neq \det(B)$, and hence $A$ and $B$ are not similar.

###
(b) Is the matrix $A=\begin{bmatrix}

0 & 1\\

5& 3

\end{bmatrix}$ similar to the matrix $B=\begin{bmatrix}

1 & 2\\

4& 3

\end{bmatrix}$?

It is straightforward to check that $\det(A)=-5=\det(B)$. Thus determinants does not help here.

We recall that if $A$ and $B$ are similar, then their traces are the same. (See Problem “Similar matrices have the same eigenvalues“.)

We compute

\begin{align*}

\tr(A)=0+3=3 \text{ and } \tr(B)=1+3=4,

\end{align*}

and thus $\tr(A)\neq\tr(B)$.

Hence $A$ and $B$ are not similar.

### (c) Is the matrix $A=\begin{bmatrix}

-1 & 6\\

-2& 6

\end{bmatrix}$ similar to the matrix $B=\begin{bmatrix}

3 & 0\\

0& 2

\end{bmatrix}$?

We see that

\[\det(A)=6=\det(B) \text{ and } \tr(A)=5=\tr(B).\]
Thus, the determinants and traces do not give any information about similarity.

The characteristic polynomial of $A$ is given by

\begin{align*}

p(t)&=\det(A-tI)\\

&=\begin{vmatrix}

-1-t & 6\\

-2& 6-t

\end{vmatrix}\\

&=(-1-t)(6-t)-(6)(-2)\\

&=t^2-5t+6.

\end{align*}

(Note that since we found the determinant and trace of $A$, we could have found the characteristic polynomial from the formula $p(t)=t^2-\tr(A)t+\det(A)$.)

Since $p(t)=(t-2)(t-3)$, the eigenvalue of $A$ are $2$ and $3$.

Since $A$ has two distinct eigenvalues, it is diagonalizable.

That is, there exists a nonsingular matrix $S$ such that

\[S^{-1}AS=\begin{bmatrix}

2 & 0\\

0& 3

\end{bmatrix}=B.\]
Thus, $A$ and $B$ are similar.

### (d) Is the matrix $A=\begin{bmatrix}

-1 & 6\\

-2& 6

\end{bmatrix}$ similar to the matrix $B=\begin{bmatrix}

1 & 2\\

-1& 4

\end{bmatrix}$?

We see that

\[\det(A)=6=\det(B) \text{ and } \tr(A)=5=\tr(B).\]
It follows from the formula $p(t)=t^2-\tr(A)t+\det(A)$ (or just computing directly) that the characteristic polynomials of $A$ and $B$ are both

\[t^2-5t+6=(t-2)(t-3).\]
Thus, the eigenvalues of $A$ and $B$ are $2, 3$. Hence both $A$ and $B$ are diagonalizable.

There exist nonsingular matrices $S$ and $P$ such that

\[S^{-1}AS=\begin{bmatrix}

2 & 0\\

0& 3

\end{bmatrix} \text{ and } P^{-1}BP=\begin{bmatrix} 2 & 0\\

0& 3

\end{bmatrix}. \]
So we have $S^{-1}AS=P^{-1}BP$, and hence

\[PS^{-1}ASP^{-1}=B.\]
Putting $U=SP^{-1}$, we have

\[U^{-1}AU=B.\]
(Since the product of invertible matrices is invertible, the matrix $U$ is invertible.)

Therefore $A$ and $B$ are similar.

## Related Question.

For more problems about similar matrices, check out the following posts:

- Problems and solutions about similar matrices
- A matrix similar to a diagonalizable matrix is also diagonalizable

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## 1 Response

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