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

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 $$b=c$$.

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

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 $$Q^{-1} = \begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix}$$

• Those are the same as what I had originally gotten. Thanks for taking the time. It seems I simply miscalculated the RHS. May 11, 2023 at 2:16
• @user129393192 suggest you correct the numerous errors in your post above. May 11, 2023 at 2:46

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.

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.

• I do not have that theorem. Could you please show it using the definition? May 11, 2023 at 1:54
• Have you learned how to diagonalize a matrix? May 11, 2023 at 1:57
• Not in the context of this course. We do not have the definition. I am in my second course of linear algebra, where it is all proof based. May 11, 2023 at 1:57
• Maybe this helps: en.wikipedia.org/wiki/Diagonalizable_matrix May 11, 2023 at 2:01
• This post shows the proof: math.stackexchange.com/questions/1577445/…. @user129393192 May 11, 2023 at 2:08