Lecture 25 - Eigenvalues (cont.)

We continue to prove the following:

Theorem

Suppose $T \in L (V)$ and there is a basis $v_{1}, . . ., v_{n}$ denoted $β$ , for which $M (T, β)$ is U.T. Then $T$ is invertible iff every diagonal entry in $M (T, β)$ is non-zero.

Proof
Consider $\to$ . We have:

M (T) = [\begin{matrix} * & \dots \\ 0 & * & \dots \\ ⋮ & 0 & * & \dots \end{matrix}]

Assume $a_{k k}$ entry of this matrix is 0:

M (T) = [\begin{matrix} * & \dots \\ 0 & 0 & \dots \\ ⋮ & 0 & * \end{matrix}]

The bottom right matrix is still the zero matrix. Notice that $T (v_{1}), . . ., T (v_{k})$ is LI while they are in the $span (v_{1}, . . ., v_{k})$ . Which is a contradiction. See the proof at Lecture 24 - Finishing Eigenstuff#^d13947 for a more specific example.

Now consider $\leftarrow$ . Suppose the diagonal entries of $M (T)$ are all non-zero:

M (T) = [\begin{matrix} λ_{1} & \dots \\ 0 & λ_{2} & \dots \\ 0 & 0 & ⋱ \\ ⋮ & 0 & 0 & λ_{n} \end{matrix}]

where $λ_{i}$ is non-zero for all $1 \leq i \leq n$ . We need to show $T$ is invertible. Since $T$ is from $V$ to $V$ , then we only need to prove either injectivity or surjectivity.

We will show $T$ is surjective. The first column says that:

T (v_{1}) = λ_{1} v_{1} + 0 v_{2} + \dots = λ_{1} v_{1}

Since $λ_{1} \neq 0$ then:

v_{1} = \frac{T (v_{1})}{λ_{1}} = T (\frac{1}{λ} v_{1})

So then $v_{1} \in range (T)$ .

Repeat for the second column. Here:

T (v_{2}) = a_{12} v_{1} + λ_{2} v_{2} \Rightarrow \frac{T (v_{2}) - a_{12} v_{1}}{λ_{2}} = v_{2} = T (\frac{v_{2}}{λ_{2}}) - \frac{a_{12}}{λ_{2}} v_{1}

Thus $v_{2} \in range (T)$ since the right vector is already in the range of $T$ .

Repeat this process, up to $n$ . Hence, all $v_{i} \in range (T)$ , so then $V = range (T)$ showing $T$ is surjective.
☐

We're now going to use this theorem to prove a bigger theorem:

Theorem

Suppose $T \in L (V)$ that has an U.T. matrix representation with respect to the basis $v_{1}, . . ., v_{n}$ , denoted $β$ . Then the eigenvalues of $T$ are precisely the diagonal entries of $M (T, β)$ .

Proof
Notice that $λ$ is an eigenvalue of $T$ iff $T - λ I$ is not invertible via Lecture 22 (online) - Invariant Subspaces#^5fc8ac. Thus, $M (T - λ I)$ has a zero on the diagonal. Note that:

(T - λ I) v_{i} = T (v_{i}) - λ v_{i}

And the matrix in a similar way:

M (T - λ I) = M (T) - λ I_{n \times n}

So if:

M (T) = [\begin{matrix} λ_{1} & \dots \\ 0 & ⋱ & \dots \\ 0 & \dots & λ_{n} \end{matrix}]

Then we get the new matrix is:

M (T - λ I) = M (T) - λ I_{n \times n} = [\begin{matrix} λ_{1} - λ & \dots & * \\ 0 & ⋱ & \dots \\ 0 & \dots & λ_{n} - λ \end{matrix}]

Thus $λ = λ_{i}$ for some $i$ since we have a zero on the diagonal. This applies for all entries, in an if-and only if- type proof.
☐

An Example

Let's do an example. Let $V = span (e^{x} + e^{- x}, e^{x} - e^{- x}) = span (v_{1}, v_{2})$ . Note that $v_{1}, v_{2}$ is a basis $β$ for $V$ so then $dim (V) = 2$ . Let $D$ denote the derivative map. Let's look at:

M (D, β) = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}]

since:

D (v_{1}) = D (e^{x} + e^{- x}) = e^{x} - e^{- x} = v_{2}

and:

D (v_{2}) = D (e^{x} - e^{- x}) = e^{x} + e^{- x} = v_{1} $ $ N o t i c e t h a t $ M (D, β) $ i s n^{'} t u p p e r t r i a n g u l a r . B u t w h a t i f w e w a n t t o c h a n g e o u r b a s i s v e c t o r s t o m a k e i t u p p e r t r i a n g u l a r ? N o t e t h a t :

v_1 + v_2 = 2e^x, D(v_1 + v_2) = 2e^x

t h i s i s a n e i g e n v e c t o r f o r $ D $ . T h u s, $ v_{1} + v_{2} = v_{λ} $ i s a n e i g e n v e c t o r o f e i g e n v a l u e $ λ = 1 $ . L e t^{'} s c r e a t e a n e w b a s i s $ β^{'} $ u s i n g $ v_{1} + v_{2} $ a n d a n e w v e c t o r $ v_{λ}^{'} = 4 e^{x} - 3 e^{- x} $ . T h e n :

M(D, \beta') = \begin{bmatrix}
1 & 4 \
0 & -1
\end

t h e f i r s t c o l u m n i s t r i v i a l, a n d f o r t h e s e c o n d c o l u m n :

D(4e^x - 3e^{-x}) = 4e^x + 3e^x = 4(2e^x) - (4e^x - 3e^{-x}) = 4v_\lambda - v_\lambda'

So since $M(D,\beta')$ has no zero then $D$ is invertible (there's no zeros on the diagonal). Furthermore, $\lambda = 1$ and $\lambda = -1$ are the eigenvalues of $D$. # Diagonal Matrix Representations Recall that: > [!definition] Diagonal Matrix > $A_{n \times n}$ is called *diagonal* if all entries off the diagonal are 0's. For instance, $\text{diag}(1,2,3) = \begin{bmatrix}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3\end{bmatrix}$. The zero matrix is a diagonal matrix. Note that diagonal matrices are always U.T., so then all the properties we got from them apply, such as reading the eigenvalues off the diagonal. But we want to ask **when** a diagonal matrix $M(T)$ for $T \in \mathcal{L}(V)$ exists? Suppose:

M(T) = \begin{bmatrix}
\lambda_1 & 0 & \dots \
0 & \lambda_2 & \dots \
0 & 0 & \ddots & \
0 & 0 & \dots & \lambda_n
\end

T h e n :

\begin{align}
T(v_1) = \lambda_1 v_1 \
T(v_2) = \lambda_2 v_2 \
\vdots \
T(v_n) = \lambda_n v_n
\end

s o a l l $ v_{i} $ a r e e i g e n v e c t o r s w i t h a s s o c i a t e d e i g e n v a l u e s $ λ_{i} $ . T h i s i s a c t u a l l y a n * e i g e n b a s i s *, n a m e l y j u s t a b a s i s c o n s i s t i n g o f e i g e n v e c t o r s . B u t t h i s w o r k s b o t h w a y s . I f y o u s t a r t e d w i t h t h e e i g e n v e c t o r s y o u c a n c o n s t r u c t t h e m a t r i x b a s e d o n t h e e i g e n b a s i s . T h i s g i v e s r i s e t o d e f i n e a n e i g e n s p a c e :> [! d e f i n i t i o n] E i g e n s p a c e > G i v e n $ T \in L (V) $, $ λ $ i s a n e i g e n v a l u e o f $ T $, t h e n :>

E(\lambda, T) = \text{null}(T - \lambda I)
$i s t h e s e t o f a l l e i g e n v e c t o r s c o r r e s p o n d i n g t o $ λ $ a l o n g w i t h $ \vec{0} $ . T h i s i s t h e * e i g e n s p a c e * a s s o c i a t e d w i t h $ λ $ .$

Some facts about eigenspaces:

$E (λ, T)$ is $T$ invariant.
If $λ_{1}, . . ., λ_{m}$ are distinct, then $\sum_{i = 1}^{m} E (λ_{i}, T)$ is a direct sum.