0
$\begingroup$

Also how do I show that a matrix is not diagonalizable based on my calculations from the equation $\det(A-\lambda I)=0$

For the first part say the matrix is:

$$ \begin{pmatrix} 1 & -2 & 3 \\ 0 & -1 & 3 \\ -1 & 2 & -2 \\ \end{pmatrix} $$ is $\lambda=1$ an eigenvalue for the matrix?

Also I wouldn't mind finding out the code for lambda on here and the code for putting a set vectors in brackets as related to span.

Thanks!

$\endgroup$
  • $\begingroup$ Code for lambda is "\lambda" between dollar signs. $\endgroup$ – Dave May 1 '16 at 20:43
  • $\begingroup$ Also, the curly brackets { and } have to be "escaped" in MathJax/$\LaTeX$ markup by preceding them with backslash `\`. See the MathJax Basic Tutorial and Quick Reference for many of the common techniques. $\endgroup$ – hardmath May 19 '16 at 12:16
1
$\begingroup$

To find the eigenvalues of $A=\begin{bmatrix} 1&-2&3\\0&-1&3\\-1&2&-2\end{bmatrix}$ we compute the roots of the characteristic polynomial, $p_A(t)=det(tI-A)$. Using Laplace Expansion we have: $$p_A(t)=det\begin{bmatrix} t-1&2&-3\\0&t+1&-3\\1&-2&t+2\end{bmatrix}=(t-1)det\begin{bmatrix} t+1&-3\\-2&t+2 \end{bmatrix}+det\begin{bmatrix} 2&-3\\t+1&-3 \end{bmatrix}=(t-1)[(t+1)(t+2)-6]+[-6+3t+3]=(t-1)(t^2+3t-1)$$ From factored form, we see that the eigenvalues of $A$ are $\lambda_1=1, \lambda_2=\frac{-3+\sqrt {13}}{2}, \lambda_3=\frac{-3-\sqrt {13}}{2}$. So to answer the first question: yes, the matrix has $1$ as an eigenvalue. Now, because the matrix is $3\times 3$ and it has $3$ distinct eigenvalues, it is diagonalizable.

Note that showing an $n\times n$ matrix is diagonalizable is equivalent to showing that the algebraic multiplicities of its eigenvalues equals their geometric multiplicities. In the special case where it has $n$ distinct eigenvalues, this condition is immediately seen to hold true, and the matrix is diagonalizable.

$\endgroup$
-1
$\begingroup$

After subtracting the eigenvalues $\lambda$ from diagonal elements are some rows or columns or after some linear combinations the same? Does it vanish?

$$ \begin{pmatrix} 0 & -2 & 3 \\ 0 & -2 & 3 \\ -1 & 2 & -3 \\ \end{pmatrix} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.