Use the definition of the matrix exponential to compute $e^A$ for $A$

Question

Use the definition of the matrix exponential to compute $e^A$ for $A=\begin{pmatrix} 1 & 1\\ 0 &1 \end{pmatrix}$

My work:

I know that $e^A=Xe^DX^{-1}$

To obtain $e^D$, I calculated for the eigen values $\lambda_1 = 1$ and $\lambda_2 = 1$

Therefore $e^D= \begin{pmatrix} e^1 & 0\\ 0 &e^1 \end{pmatrix}$

For the eigen values, I got the eigen vectors: x$_1 = (1,0)^T$ and x$_2 = (1,0)^T$

This means that $X= \begin{pmatrix} 1 & 1\\ 0 &0 \end{pmatrix}$

Then, I tried to find $X^{-1}$, but my $X$ is not invertible and I really don't know where I made my mistake.

Answer 1

Your $ X $ is not invertible. More importantly, $ A $ is not diagonalizable otherwise since its only eigenvalue is $1 $ it would be equal to the identity.

With your work, you found another proof of the non diagonalizability, since the geometric multiplicity ($1$) is less than the algebraic multiplicity ($2$) for the eigenvalue $1 $.

Hint: use induction to prove a formula for $ A^n $.

Answer 2

Hint The question asks for use of the definition. The definition you are looking for is:

$$e^A = \sum_{k=0}^{\infty} \frac{1}{k!} A^k$$

First try computing $A^0$, $A^1$, $\ldots$, you should notice a pattern and be able to prove a formula using induction for this. After this I would attempt to evaluate $e^A$.

Answer 3

Hint:

It is useful to write $A= I +N$ with $I$ the identity matrix and $$N=\begin{pmatrix}0&1\\0 &0\end{pmatrix}.$$ Note that $I$ and $N$ commute and thus $\exp(A) =\exp(I) \exp(N)$. Now $\exp(I)$ is trivial and so is $\exp(N)$ (as $N^2=0$).

Answer 4

Write $A$ as $I_2 + B$ where $B = \left[\begin{smallmatrix}0 & 1\\0 & 0\end{smallmatrix}\right]$. Since $B^2 = 0$, we have

$$A^n = (I_2+B)^n = I_2 + \binom{n}{1} B + \binom{n}{2}B^2 + \cdots = I_2 + nB$$

By definition of $e^A$, we get $$e^A = \sum_{n=0}^\infty \frac{A^n}{n!} = \sum_{n=0}^\infty \frac{I_2+nB}{n!} = \left(\sum_{n=0}^\infty \frac{1}{n!} \right) I_2 + \left(\sum_{n=1}^\infty \frac{1}{(n-1)!}\right) B = e(I_2+B) = eA$$