1. matrix exponential

$\displaystyle e^A=Xe^DX^{-1}$

$\displaystyle A=\begin{bmatrix} 1 & 1\\ -1 & -1 \end{bmatrix}$

$\displaystyle \lambda_1=\lambda_2=0$

This will yield only 1 eigenvector so this matrix can't be diagonlaized.

How can I do this problem?

2. Hello,

Actually, $\displaystyle e^A=\sum_{k=0}^\infty \frac{1}{k!} \cdot A^k=I+A+\sum_{k=2}^\infty \frac{1}{k!} \cdot A^k$

but if you look at A, you'll see that $\displaystyle A^2=\begin{pmatrix}0&0\\0&0\end{pmatrix}$

so $\displaystyle A^k=\begin{pmatrix}0&0\\0&0\end{pmatrix} ~,~ \forall k\geq 2$

finally, $\displaystyle e^A=I+A$

3. Is there a way to do this with diagonal matrices like first wrote?

4. No.

And I showed you this because the method of diagonalizing comes from this formula.

If one can diagonalize the matrix A, then we have $\displaystyle A=XDX^{-1}$, where D is a diagonal matrix.
So $\displaystyle A^k=XD^kX^{-1}$, and $\displaystyle D^k$ is always easy to compute. That's why it is set this way almost every time you deal with the exponential of a matrix.

If you want to see how it can be used with a diagonalizable matrix, tell me, I'll find a previous thread in this forum for ya

5. Then neither of these matrices can be done via diagonalization:

$\displaystyle \begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix}$ and $\displaystyle \begin{bmatrix} 1 & 0 & -1\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$ then either, correct?

6. If I'm not mistaking, they're both diagonalizable, so it can be done with the method you want so much to use

(for both of them, the corresponding diagonal matrix should be the identity matrix)

7. Originally Posted by Moo
If I'm not mistaking, they're both diagonalizable, so it can be done with the method you want so much to use

(for both of them, the corresponding diagonal matrix should be the identity matrix)
When I did the diagonalizing, the eigenspaces were less than n.

8. For the first one, 1 is an eigenvalue with multiplicity 2, and for the second one, 1 is an eigenvalue with multiplicity 3.
The eigenspace related to 1 is respectively of dimension 2 and 3.
There's no doubt about it, use characteristic polynomials

9. $\displaystyle \begin{bmatrix} 1-\lambda & 1\\ 0 & 1-\lambda \end{bmatrix}\Rightarrow\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}\Rightarrow x_1\begin{bmatrix} 1 \\ 0 \end{bmatrix}$

I only have one egienvector. What went wrong?

10. Originally Posted by Moo
If I'm not mistaking, they're both diagonalizable, so it can be done with the method you want so much to use

(for both of them, the corresponding diagonal matrix should be the identity matrix)
No, neither is diagonalizable. The first is already in "Jordan Normal Form" and the other can be put in that form.

As you say, "1" is the only eigenvalue and <1, 0, 0> and <0, 1, 0> span the two-dimensional eigenspace.

Ir can be put in Jordan Normal Form $\displaystyle N= \begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$ which would make it slightly easier to find powers or the exponential of the matrix (though not as easy as if it were diagonal). For example,
$\displaystyle N^2= \begin{bmatrix}1 & 2 & 0 \\ 0 & 1 & 0 \\0 & 0 & 1\end{bmatrix}$
$\displaystyle N^3= \begin{bmatrix}1 & 3 & 0 \\ 0 & 1 & 0 \\0 & 0 & 1\end{bmatrix}$
$\displaystyle N^4= \begin{bmatrix}1 & 4 & 0 \\ 0 & 1 & 0 \\0 & 0 & 1\end{bmatrix}$
and, in general,
$\displaystyle N^k= \begin{bmatrix}1 & k & 0 \\ 0 & 1 & 0 \\0 & 0 & 1\end{bmatrix}$

$\displaystyle e^N= $$\displaystyle \begin{bmatrix}1+ 1+ 1/2+ 1/6+...+ 1/k!+ ...& 1+ 1+ 2/2+ 3/6+ ...k/k!+ ... & 0\end{bmatrix}$$\displaystyle \begin{bmatrix} 0 & 1+ 1+ 1/2+ 1/6+ ...+ 1/k!+... & 0 \\ 0 & 0 & 1+ 1+ 1/2+ 1/6+ ...+ 1/k!+ ...\end{bmatrix}$
(Sorry about the break in the matrix, but the original was too large to fit the LaTex.)

Note that 1+ 1+ 2/2+ 3/6+ ...+ k/k!+ ...= 1+ (1+ 1+ 1/2+ ...+ 1/(k-1)!+ ... = 1+ e.

That is, $\displaystyle e^N= \begin{bmatrix}e & 1+ e & 0 \\ 0 & e & 0\\ 0 & 0 & e\end{bmatrix}$.

11. Owww... Sorry I promise I won't reply to any linear/abstract algebra for a long time ! (not a big loss )