# Diagonalizing 2x2 matrices

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• June 21st 2010, 06:17 PM
Cursed
Diagonalizing 2x2 matrices
Diagonalize the following matrix:

$
\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:

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

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

For $\lambda = 5$:

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

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

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

____________________

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

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

so

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

__________________________

Now:

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

^ Is that wrong?
• June 21st 2010, 06:28 PM
maddas
Doesn't look too diagonal to me (:
• June 21st 2010, 06:45 PM
tonio
Quote:

Originally Posted by Cursed
Diagonalize the following matrix:

$
\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:

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

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

For $\lambda = 5$:

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

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

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

____________________

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

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

so

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

__________________________

Now:

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

^ Is that wrong?

.

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

Tonio
• June 21st 2010, 07:09 PM
dwsmith
Characteristic polynomials of 2x2 matrices can be done without computing $det(A-\lambda I)$
$P_{\lambda}(A)=\lambda^2-\lambda tr(A)+det(A)$

Thus, $tr(A)=3 \ \mbox{and} \ det(A)=-10\rightarrow \ \lambda^2-3\lambda-10=(\lambda-5)(\lambda+2)$
• June 22nd 2010, 02:45 AM
Ackbeet
I agree with Tonio that the $v_{2}$ eigenvector is incorrect. Your steps are good up until the moment you say that $v_{2}=\dots$. Don't skip steps there!

As for constructing $P$, you may need to run the Gram-Schmidt procedure on your eigenvectors. $P$ 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?
• June 22nd 2010, 03:47 AM
Defunkt
Perhaps I'm being confused, but why would P need to be orthogonal?
• June 22nd 2010, 04:39 AM
Ackbeet
Quote:

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 $P$ is as simple as transposition. You are correct (implying I was wrong before) in saying that $P$ does not have to be orthogonal. It does have to be invertible.