# Some problem on eigenvalues,eigenvectors and diagonalizability

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• Nov 30th 2011, 08:19 AM
maoro
Some problem on eigenvalues,eigenvectors and diagonalizability
I'm preparing final exam on linear algebra and come across with these problem:

#1
Let A be an n*n matrix.Assume that for some $k>=1$, $A^k=/=0$ and $A^kA=0$.
a)Find all the eigenvalues of A.
b)Prove that A is not diagonalizable.
c)Prove that k<n
d)If k=n-1,find the dimension of null A

#2
Let A be an n*n matrix and $a=/=b$ be two real numbers.Assume that
$(A-aI)(A-bI)=0$
Prove that A is diagonalizable.

#3
Let $k>=2$ and $X_1,X_2,...,X_n$ be eigenvectors correspond to distinct eigenvalues $n_1,n_2,...,n_k$ of an n*n matrix A.
Prove that $X_1+X_2+...+X_k$ is not an eigenvector of A.

For #3,
Suppose on the contrary, $X_1+X_2+...+X_k$ is an eigenvector of A
Then there exists a number n such that
$A(X_1+X_2+...+X_k)=n(X_1+X_2+...+X_k)$
$=> AX_1+AX_2+...+AX_n=n(X_1+X_2+...+X_k)$
$=> n_1X_1+n_2X_2+...+n_kX_k=n(X_1+X_2+...+X_k)$

$=>(n_1-n)X1+(n_2-n)X2+...+(n_k-n)X_k=0$

Since eigenvectors are lin. ind.
therefore, $(n_1-n)=(n_2-n)=...=(n_k-n)=0 => n_1=n_2=...=n_k=n$
Contradict to $n_1,n_2,...,n_k$are distinct.

Is it right ?
• Nov 30th 2011, 10:22 AM
FernandoRevilla
Re: Some problem on eigenvalues,eigenvectors and diagonalizability
Quote:

Originally Posted by maoro
Let A be an n*n matrix.Assume that for some $k>=1$, $A^k=/=0$ and $A^kA=0$.
a)Find all the eigenvalues of A.

If $\lanbda$ is an eigenvalue of $A$ there exists $0\neq x\in\mathbb{R}^n$ such that $Ax=\lambda x$ . Then, $0=A^{k+1}x=\ldots=\lambda^{k+1}x\;(x\neq 0)$ which implies $\lambda=0$

Quote:

b)Prove that A is not diagonalizable.
$A^{k+1}=0\Rightarrow (\det A)^{k+1}=0\Rightarrow \det A=0\Rightarrow \textrm{rank}A etc ...

Try the rest of problem #1

P.S. Please, see rule #8 here:

http://www.mathhelpforum.com/math-he...hp?do=vsarules
• Nov 30th 2011, 10:40 AM
FernandoRevilla
Re: Some problem on eigenvalues,eigenvectors and diagonalizability
Quote:

Originally Posted by maoro
#2Let A be an n*n matrix and $a=/=b$ be two real numbers.Assume that $(A-aI)(A-bI)=0$ Prove that A is diagonalizable.

One way: $p(x)=(x-a)(x-b)$ is an annihilator polynomial of $A$ so the possible minimal polynomials of $A$ are $\mu(x)=x-a$ , $\mu(x)=x-b$ or $\mu(x)=(x-a)(x-b)$ . In all cases, the blocks of the canonical form of $A$ have dimension $1$, so $A$ is diagonalizable.

P.S. Comment us if you have not covered Jordan canonical forms.
• Nov 30th 2011, 06:55 PM
maoro
Re: Some problem on eigenvalues,eigenvectors and diagonalizability
Sorry for posting too many questions on a single thread.

btw, I have not covered Jordan canonical forms and so is there any alternative approach for #2 ?
• Nov 30th 2011, 08:29 PM
FernandoRevilla
Re: Some problem on eigenvalues,eigenvectors and diagonalizability
Quote:

Originally Posted by maoro
is there any alternative approach for #2 ?

Yes, there is. Hint : Consider $F_1=\textrm{Im}(A-aI),\;F_2=\textrm{Im}(A-bI)$ . Prove that $F_1\subset \ker (A-bI)$ , $F_2\subset \ker (A-aI)$ and $\mathbb{R}^n=F_1\oplus F_2$ .