# Prove that matrices $\tiny\begin{pmatrix} 2&-1 \\ 0&2 \\ \end{pmatrix},\begin{pmatrix} 2& 0 \\ 1&2 \\ \end{pmatrix}$ are similar. Error in my method?

Show that the matrix

$$\begin{pmatrix} 2&-1 \\ 0&2 \\ \end{pmatrix}$$ is similar to a triangular matrix of the form $$\begin{pmatrix} \lambda& 0 \\ 1&\lambda \\ \end{pmatrix}$$ where $\lambda$ is an eigen value.

Attempt:

The eigenvalues of the matrix $\begin{pmatrix} 2&-1 \\ 0&2 \\ \end{pmatrix}$ are clearly $2,2$.

Hence, we need to show that the matrix $\begin{pmatrix} 2&-1 \\0&2 \\ \end{pmatrix}$ is similar to $\begin{pmatrix}2&0 \\1&2 \\\end{pmatrix}$

Now, we know that two matrices are similar if and only if they represent the same linear transformation.

But: for any two dimensional vector $(x ~~y)^T :$

$\begin{pmatrix} 2&-1 \\ 0&2 \\ \end{pmatrix} \begin{pmatrix}x \\y \\ \end{pmatrix} = \begin{pmatrix} 2x-y \\ 2y \\ \end{pmatrix}$

And

$\begin{pmatrix} 2&0 \\1&2 \\\end{pmatrix} \begin{pmatrix}x \\ y \\\end{pmatrix} = \begin{pmatrix} 2x \\x+2y \\\end{pmatrix}$

Clearly, $\begin{pmatrix} 2x \\x+2y \\\end{pmatrix}$ and $\begin{pmatrix} 2x-y \\2y \\ \end{pmatrix}$ don't represent the same linear transformation.

So, how can these matrices be similar?

What could be the fault in my reasoning?

Thank you very much for your help in this regard.

• Why is that so clear? Remember, the idea of similarity is that they represent the same linear transformation but in different bases. – John Brevik Apr 7 '15 at 17:34
• iwont it be easier to use the fact that $A$ and $B$ are similar iff $A - kI$ and $B - kI$ are similar. in your case take $k = 2.$ – abel Apr 7 '15 at 17:40
• @JohnBrevik If $B = C^{-1}AC$, then under the same basis, the matrix $C$ will be simply the identity matrix. Could you please explain why it's necessary for the two bases to be different? Thanks. – MathMan Apr 7 '15 at 17:52
• OK, I should have said "possibly different bases." – John Brevik Apr 7 '15 at 19:01
• @JohnBrevik Uhm, I guess I was a little confused. But, I think I get it now. Thank you very much for your comment. :) – MathMan Apr 7 '15 at 19:05

## 3 Answers

Call $A$ the first matrix, written with respect to a basis $e_{1}, e_{2}$ and $B$ the second, written with respect to a basis $f_{1}, f_{2}$.

Note that $A$ fixes $e_{1}$, and $B$ fixes $f_{2}$. Moreover, $A e_{2} = 2 e_{2} - e_{1}$. Can you find a number $a$ so that for the vector $f_{1} + a f_{2}$ one has $$B (f_{1} + a f_{2}) = 2 (f_{1} + a f_{2}) - f_{2}?$$

If you can do that, the matrix of $B$ with respect to the basis $f_{2}, f_{1} + a f_{2}$ will be $A$.

• Thanks. Could you also please explain what could be wrong with my reasoning? – MathMan Apr 7 '15 at 17:56
• @Wanderer, when you say two matrices are similar that they must represent the same transformation is true. but they are with respect to different coordinate systems. you will have to do the coordinate transformation. that exactly is what the matrix $U$ does. – abel Apr 7 '15 at 18:05
• @abel If $B=C^{−1}AC$, then under the same basis, the matrix $C$ will be simply the identity matrix. So, would it be necessary for the two bases to be different? Thanks – MathMan Apr 7 '15 at 18:09
• @Wanderer, for $A, B$ to be similar means they have the same form with respect to two (different) bases, or equivalently $B = C^{-1} A C$, where $C$ is the matrix of the change of bases. To make a particularly simple example, consider the linear transformation that has matrix $A = \begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$ with respect to a basis $e_{1}, e_{2}$. You have $A(e_{1} + e_{2}) = e_{1} + e_{2}$, and $A(e_{1} - e_{2}) = - (e_{1} - e_{2})$. [follows in next comment] – Andreas Caranti Apr 7 '15 at 18:45
• So with respect to the basis $f_{1} = e_{1} + e_{2}$ and $f_{2} = e _{1} - e_{2}$, the linear transformation will have matrix $B = \begin{bmatrix}1 & 0\\0 & -1\end{bmatrix}$, and $A, B$ will be similar. Taking $C = \begin{bmatrix}1 & 1\\1 & -1\end{bmatrix}$, we will have $B = C^{-1} A C$. – Andreas Caranti Apr 7 '15 at 18:45

we can show that $A=\pmatrix{0&-1\\0&0}$ and $B= \pmatrix{0&0\\1&0}$ similar by explicitly displaying $$\pmatrix{1&0\\0&0}=AU = \pmatrix{0&-1\\0&0} \pmatrix{0&1\\-1&0} = \pmatrix{0&1\\-1&0} \pmatrix{0&0\\1&0} = UB= \pmatrix{1&0\\0&0}.$$

now, $$AU=UB \implies (A+2I)U=U(B+2I) .$$ therefore $A+2I$ and $B+2I$ are similar.

• Thanks. Could you also please explain what could be wrong with my reasoning? – MathMan Apr 7 '15 at 17:53

Let $(e_1, e_2)$ a basisi corresponding to the first matrix: $$Ae_1= 2e_1 \\ Ae_2= - e_1 + 2e_2$$

Consider another basis $(x_1, x_2)$. We want $Ax_2 = x_2$, let us take $x_2 = e_1$. Then look for $x_2 = ae_1 + be_2$.

Can you find a value of $(a,b)$ such as the matrix in the basis $(x_1, x_2)$ is the second one?

• Thanks. I think I can find such a value. But, could you also please explain what could be wrong with my reasoning? – MathMan Apr 7 '15 at 17:56
• if $A$ and $B$ are different, then the change of basis matrix $C$ will necessarily be different from the identity matrix. – abel Apr 7 '15 at 18:15
• @abel and so what? – mookid Apr 7 '15 at 22:21
• mookid, i made a mistake by commenting from your post instead of mine. i did that twice. sorry for the inconvenience. – abel Apr 7 '15 at 22:40