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!
Best Answer
Hint Next step would be to deduce the dimension of the eigenspaces; and then use the fact that a matrix is diagonizable iff there exists a basis of eigenvectors