# Andrew's question at Yahoo! Answers (Similar matrices)

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Here is the question:

Let A =
7 -8 0
4 -5 0
-1 1 1

If possible, find a diagonal matrix C and an invertible matrix B such that A = BCB^-1

I have found the characteristic polynomial of A, along with the eigenvalues/vectors. However I just don't understand this question, and my notes/textbook is completely useless! I assume the two matrices A and C will be similar, thus have the same eigenvalues. But how am I supposed to show that matrix!?
Here is a link to the question:

Question on Similar Matrices? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Hello Andrew,

The characteristic polynomial of $A$ is: \begin{aligned}\chi (\lambda)&=\begin{vmatrix}{7-\lambda}&{-8}&{0}\\{4}&{-5-\lambda}&{0}\\{-1}&{1}&{1-\lambda}\end{vmatrix}\\&=(1-\lambda)\begin{vmatrix}{7-\lambda}&{-8}\\{4}&{-5-\lambda}\end{vmatrix}\\&=(1-\lambda)(\lambda+1)(\lambda-3)\end{aligned} The eigenvalues are $1,-1,3$ (all simple) and according to a well-known property, the matrix is diagonalizable. The eigenspaces (with corresponding basis) are: $$V_1\equiv\left \{ \begin{matrix}6x_1-8x_2=0\\4x_1-6x_2=0\\-x_1+x_2=0\end{matrix}\right.\qquad B_{V_1}=\{(0,0,1)\}$$ $$V_{-1}\equiv\left \{ \begin{matrix}8x_1-8x_2=0\\4x_1-4x_2=0\\-x_1-x_2+2x_3=0\end{matrix}\right.\qquad B_{V_{-1}}=\{(1,1,1)\}$$ $$V_3\equiv\left \{ \begin{matrix}4x_1-8x_2=0\\4x_1-8x_2=0\\-x_1-x_2-2x_3=0\end{matrix}\right.\qquad B_{V_3}=\{(4,2,-1)\}$$ So, if $$B=\begin{bmatrix}{0}&{1}&{\;\;4}\\{0}&{1}&{\;\;2}\\{1}&{1}&{-1}\end{bmatrix}\;,\quad C=\begin{bmatrix}{1}&{\;\;0}&{0}\\{0}&{-1}&{0}\\{0}&{\;\;0}&{3}\end{bmatrix}$$ then, $A=BCB^{-1}$. You can easily verify this equality proving the equivalent one $AB=BC$.