1. ## Diagonalizing 2x2 matrices

Diagonalize the following matrix:

$\displaystyle \begin{bmatrix} 2 & 3 \\ 4 &1 \\ \end{bmatrix}$

I don't know if I made an algebraic mistake or if the matrix is just not diagonalizable. o.O

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My work:

$\displaystyle det(A-\lambda I) = det\begin{bmatrix} 2-\lambda & 3 \\ 4 &1-\lambda \\ \end{bmatrix} = (2-\lambda)(1-\lambda)-(3)(4) =(\lambda-5)(\lambda+2)$

Therefore: $\displaystyle \lambda = 5, \lambda = -2$
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For $\displaystyle \lambda = 5$:

$\displaystyle \begin{bmatrix} 2 & 3 \\ 4 & 1 \\ \end{bmatrix} - \begin{bmatrix} 5 & 0 \\ 0 & 5 \\ \end{bmatrix} = \begin{bmatrix} -3 & 3 \\ 4 & -4 \\ \end{bmatrix}$ ~ $\displaystyle \begin{bmatrix} 1 & -1 \\ 0 & 0 \\ \end{bmatrix}$

$\displaystyle (A - 5\lambda)X = 0$: $\displaystyle x_1 -x_2 = 0$ and $\displaystyle x_2$ is free. So $\displaystyle x_1 = x_2$, but $\displaystyle x_2 =1$, which means $\displaystyle x_1 = 1$ too.

$\displaystyle v_1= \begin{bmatrix}1 \\ 1 \\ \end{bmatrix}$
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Using the same method for $\displaystyle \lambda = -2$, I get:

$\displaystyle (A+2I)X= \begin{bmatrix} 4 & 3 \\ 4 & 3 \\ \end{bmatrix}$ ~ $\displaystyle \begin{bmatrix} 4 & 3 \\ 0 & 0 \\ \end{bmatrix}$

so

$\displaystyle v_2= \begin{bmatrix}3 \\ 4 \\ \end{bmatrix}$
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Now:

$\displaystyle P= [v_1$ $\displaystyle v_2]$ $\displaystyle = \begin{bmatrix}1 & 3 \\ 1 & 4 \\ \end{bmatrix}$

^ Is that wrong?

2. Doesn't look too diagonal to me (:

3. Originally Posted by Cursed
Diagonalize the following matrix:

$\displaystyle \begin{bmatrix} 2 & 3 \\ 4 &1 \\ \end{bmatrix}$

I don't know if I made an algebraic mistake or if the matrix is just not diagonalizable. o.O

_______________________________________
My work:

$\displaystyle det(A-\lambda I) = det\begin{bmatrix} 2-\lambda & 3 \\ 4 &1-\lambda \\ \end{bmatrix} = (2-\lambda)(1-\lambda)-(3)(4) =(\lambda-5)(\lambda+2)$

Therefore: $\displaystyle \lambda = 5, \lambda = -2$
____________________

For $\displaystyle \lambda = 5$:

$\displaystyle \begin{bmatrix} 2 & 3 \\ 4 & 1 \\ \end{bmatrix} - \begin{bmatrix} 5 & 0 \\ 0 & 5 \\ \end{bmatrix} = \begin{bmatrix} -3 & 3 \\ 4 & -4 \\ \end{bmatrix}$ ~ $\displaystyle \begin{bmatrix} 1 & -1 \\ 0 & 0 \\ \end{bmatrix}$

$\displaystyle (A - 5\lambda)X = 0$: $\displaystyle x_1 -x_2 = 0$ and $\displaystyle x_2$ is free. So $\displaystyle x_1 = x_2$, but $\displaystyle x_2 =1$, which means $\displaystyle x_1 = 1$ too.

$\displaystyle v_1= \begin{bmatrix}1 \\ 1 \\ \end{bmatrix}$
____________________

Using the same method for $\displaystyle \lambda = -2$, I get:

$\displaystyle (A+2I)X= \begin{bmatrix} 4 & 3 \\ 4 & 3 \\ \end{bmatrix}$ ~ $\displaystyle \begin{bmatrix} 4 & 3 \\ 0 & 0 \\ \end{bmatrix}$

so

$\displaystyle v_2= \begin{bmatrix}3 \\ 4 \\ \end{bmatrix}$
__________________________

Now:

$\displaystyle P= [v_1$ $\displaystyle v_2]$ $\displaystyle = \begin{bmatrix}1 & 3 \\ 1 & 4 \\ \end{bmatrix}$

^ Is that wrong?
.

Yes, it is wrong: $\displaystyle \begin{pmatrix}2&3\\4&1\end{pmatrix}\begin{pmatrix }3\\4\end{pmatrix}\neq(-2)\begin{pmatrix}3\\4\end{pmatrix}$ , so that isn't an eigenvector corresponding to the eigenvalue -2...

Tonio

4. Characteristic polynomials of 2x2 matrices can be done without computing $\displaystyle det(A-\lambda I)$
$\displaystyle P_{\lambda}(A)=\lambda^2-\lambda tr(A)+det(A)$

Thus, $\displaystyle tr(A)=3 \ \mbox{and} \ det(A)=-10\rightarrow \ \lambda^2-3\lambda-10=(\lambda-5)(\lambda+2)$

5. I agree with Tonio that the $\displaystyle v_{2}$ eigenvector is incorrect. Your steps are good up until the moment you say that $\displaystyle v_{2}=\dots$. Don't skip steps there!

As for constructing $\displaystyle P$, you may need to run the Gram-Schmidt procedure on your eigenvectors. $\displaystyle P$ has to be orthogonal, which means that the eigenvectors have to be orthonormal. Also, don't forget that the final result has to be your diagonal matrix. Remember how to get the diagonal matrix?

6. Perhaps I'm being confused, but why would P need to be orthogonal?

7. Originally Posted by Defunkt
Perhaps I'm being confused, but why would P need to be orthogonal?
Maybe I'm thinking of orthogonally diagonalizing a matrix. The advantage there is that inverting the $\displaystyle P$ is as simple as transposition. You are correct (implying I was wrong before) in saying that $\displaystyle P$ does not have to be orthogonal. It does have to be invertible.