Are $\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}$ and $\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}$ similar?

Question

I have two matrices:

$$ A =\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}~~~~~~~ B = \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} $$ I am to show they are similar or not similar. I set the following up:

$$ B = Q^{-1}AQ $$ where $$ Q = \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$ I solved that $d = -b$ and $c=b$ and $-b(b-a) \ne 0$. I set $b = 2$, $a =1$ and attempted to solve $B=Q^{-1}AQ$, but I end up getting:

$$ \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 3 & 0 \end{bmatrix} $$ I wanted to ask about my general strategy, and if at this point, I must simply plug in different values to see what works.

Calculation:

$$ \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$ $$ \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} = \frac{1}{ad-bc} \begin{bmatrix} (a+c)d & (b+d)d \\ -c(a+c) & -c(b+d) \end{bmatrix} $$

I then solved the linear system.

Likewise, solving $AQ = QB$, I get:

$$ \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} a & b \\ a & b \end{bmatrix} = \begin{bmatrix} a & b \\ a & b \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} $$ $$ \begin{bmatrix} a+c & b+d \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} a+b & 0 \\ c+d & 0 \end{bmatrix} $$ which only gives that $b=c$.

Answer 1

$$ A =\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} $$ $$ B = \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} $$

where $$ Q = \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

THEN $$ AQ =\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix}= \begin{bmatrix} a & b \\ a & b \end{bmatrix} $$

WHILE

$$ QB = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}= \begin{bmatrix} a & a \\ c & c \end{bmatrix} $$

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

Finally $$ AQ -QB = \begin{bmatrix} 0 & b-a \\ a-c & b-c \end{bmatrix} $$ In order for this to be the zero matrix, we need $a=b=c.$ Once we do that, we need to have determinant nonzero to put his back into the form $Q^{-1} A Q.$ Well, now the determinant becomes $$ ad-bc \mapsto ad - a^2 = a (d-a). $$ In order for this determinant to be nonzero, we must have both $a \neq 0$ and $d \neq a$

One possibility is $$ Q = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} $$ for which $$ Q^{-1} = \begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix} $$

Answer 2

Note $A, B$ are diagonalizable with the same set of eigenvalues $\lambda_1=1, \lambda_2=0$, hence, there exists invertible matrices $P_A, P_B$, such that

$$P_A^{-1}AP_A=D=P_B^{-1}BP_B$$

where

$$D =\begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$$

Therefore,

$$A=P_AP_B^{-1}BP_BP_A^{-1}=(P_BP_A^{-1})^{-1}B(P_BP_A^{-1})$$

Let $Q=P_BP_A^{-1}$, then we have

$$A=Q^{-1}BQ$$

They are similar.

Answer 3

You can realise after some progress that the benefits of the result : $A$ and $A^T$ are similar.

Similarity have many equivalent definition:

1. $A\sim B$

2. $A, B$ represents the same linear map possibly under two different bases.

3. $A$ , $B$ have same Jordan normal form upto the permutations of jordan blocks.

You can derive a lot of interesting properties of similar matrices especially from $3$.

Checking similarity by definition quite laborious.