# Proving a matrix is diagonalizable given eigenvectors and information about characteristic polynomial ranks

Let $$A \in \Bbb R^{5 \times 5}$$. Let $$v_1=(1,0,0,1,1), \quad v_2=(1,1,0,0,1), \quad v_3=(-1,0,1,0,0)$$ be eigenvectors of $$A$$. Also, $$\rho(2I-A) \gt\rho(3I-A)$$ and $$A(1,2,2,1,3)^t=(0,4,6,2,6)^t$$ Prove that $$A$$ is diagonalizable and find a diagonal matrix similar to $$A$$.

Hint: write the vector $$u=(1,2,2,1,3)$$ as a linear combination of $$v_1,v_2,v_3$$.

What I tried:

so according to the the hint $$\alpha(1,0,0,1,1) + \beta (1,1,0,0,1) +\gamma(-1,0,1,0,0)=(1,2,2,1,3)$$ we get $$\alpha =1 , \beta = 2 \gamma = 2$$ so we have $$1\cdot(1,0,0,1,1) + 2\cdot (1,1,0,0,1) +2\cdot(-1,0,1,0,0)=(1,2,2,1,3)$$

according to $$A(1,2,2,1,3)^t=(0,4,6,2,6)^t$$ we get $$A \cdot\left(\begin{matrix} 1 \\ 2 \\ 2 \\ 1 \\ 3 \\ \end{matrix} \right)$$ $$=$$ $$\left(\begin{matrix} 0 \\ 4 \\ 6 \\ 2 \\ 6 \\ \end{matrix} \right)$$ and from the hint we get $$A \cdot\left(\begin{matrix} 1 \\ 0 \\ 0 \\ 1 \\ 1 \\ \end{matrix} \right)$$ $$+$$ $$2A \cdot\left(\begin{matrix} 1 \\ 1 \\ 0 \\ 0 \\ 1 \\ \end{matrix} \right)$$ $$+$$ $$2A \cdot\left(\begin{matrix} -1 \\ 0 \\ 1 \\ 0 \\ 0 \\ \end{matrix} \right)$$ $$=$$ $$\left(\begin{matrix} 0 \\ 4 \\ 6 \\ 2 \\ 6 \\ \end{matrix} \right)$$

According to $$Av = \lambda v$$ I got $$\lambda_1 \cdot\left(\begin{matrix} 1 \\ 0 \\ 0 \\ 1 \\ 1 \\ \end{matrix} \right)$$ $$+$$ $$2\lambda_2 \cdot\left(\begin{matrix} 1 \\ 1 \\ 0 \\ 0 \\ 1 \\ \end{matrix} \right)$$ $$+$$ $$2\lambda_3 \cdot\left(\begin{matrix} -1 \\ 0 \\ 1 \\ 0 \\ 0 \\ \end{matrix} \right)$$ $$=$$ $$\left(\begin{matrix} 0 \\ 4 \\ 6 \\ 2 \\ 6 \\ \end{matrix} \right)$$

So now I just compare and take some equations out and get that $$\lambda_1=2 , \lambda_2 =2 , \lambda_3=3$$ in order for A to be diagonalizable the values need to be different so we have atleast two eigenvalues. the eigenspaces are $$(2I-A)$$ and $$(3I-A)$$ and according to given information and dimension theorem we get

$$dimV_2 = n - \rho(2I-A) = 5- \rho(2I-A)$$ and also $$dimV_3 = n - \rho(3I-A) = 5- \rho(3I-A)$$ according to the given information of the question we can now understand that $$dimV_2 \lt dimV_3$$

eigenvectors that belong to different eigenvalues are linearly independent and we have $$\lambda_1,\lambda_2 \in V_2$$ and $$\lambda_3 \in V_3$$ so $$dimV_2 \geq 2$$ and from the conclusion that $$dimV_2 \lt dimV_3$$ we get $$dimV_3 \geq 3$$ not sure but I guess also $$dimV_3 +dimV_2 \leq 5$$ because we are in $$\Bbb R^5$$ and the dimensions can't be more than the space

This is where I got stuck.. I could not find a way to continue from to get to a matrix or prove what is related to the question.

EDIT - thanks to lhl73 , I found that $$dimV_3=3$$ and $$dimV_2=2$$ because from the equations we can get out of $$Av= \lambda v$$ we have $$\lambda_1 =(2,0,0,0,0)$$ and $$\lambda_2 = (0,2,0,0,0)$$ they are linearly independent and are in the eigenspace $$V_2$$ and we know that the sum of the dimensions can be 5 at most therefore $$dimV_3=3$$ and also according to the equations we have $$\lambda_3 = (0,0,0,3,0,0)$$ $$dimV_3=3$$ should have 2 more vectors , but how can I find them? can I just add 2 more vectors for example $$\lambda_4 = (0,0,0,0,4,0)$$ and $$\lambda_5 = (0,0,0,0,0,5)$$ that are linearly independent and then that matrix will be diagonalizable ?and the matrix made out of those 5 vectors will also be a diagonal matrix similar to A?

EDIT 2 -

I just found in the textbook that if we have $$dimV_1+...+dimV_n=dim_V$$ then the union of the basis of each dimension is the basis for $$V$$ but how can I apply this to my question? I am missing the vectors of the basis as I do not have any and guessing like in my previous edit is not actually understanding the topic so can anyone explain this to me please ? how do I find the basis

and why according to the answer from a user here, the standard basis won't work? Thanks for any help and tips!

• For notation, $\rho(\cdot)$, do you mean $\mathrm{rank}(\cdot)$? Commented Feb 4, 2022 at 5:53

• You know that there exists a basis of 2 eignevectors for $V_2$ because the dimension is 2. And similarly there exists a basis of 3 eigenvectors for $V_3$. You don't know explicitly what these vectors are, but that is not necessary: Show that for any such choice, these 2+3=5 vectors will be a basis, by showing that there can be no non-trivial linear relation among them. Commented Jan 24, 2022 at 8:39
• Yes. When the 5 eigenvectors are the columns of the matrix $P$; then $P$ is invertible (because the columns are linearly independent) and $P^{-1}AP$ is a diagonal matrix with the values $2,2,3,3,3$ on the diagonal. Commented Jan 24, 2022 at 14:53