# diagonalising a 3x3 matrix

• Aug 12th 2008, 05:55 AM
diagonalising a 3x3 matrix
im really struggling with this problem, i keep getting so far and then hitting a brick wall,
the question asks: diagonalise the matrix B=
[0 1 1]
[0 2 2]
[0 0 3]
in other words find an invertible matrix P and a diagonal matrix D so P'BP=D where P'= P inverse

• Aug 12th 2008, 06:21 AM
flyingsquirrel
Hello,
Quote:

im really struggling with this problem, i keep getting so far and then hitting a brick wall,
the question asks: diagonalise the matrix B=
[0 1 1]
[0 2 2]
[0 0 3]
in other words find an invertible matrix P and a diagonal matrix D so P'BP=D where P'= P inverse

To diagonalize $\displaystyle B$ you first need to find its eigenvalues : the roots of the polynomial $\displaystyle P(\lambda)=\det(B-\lambda I_3)$. Once you know the three eigenvalues $\displaystyle \lambda_1,\lambda_2$ and $\displaystyle \lambda_3$, you can find the eigenvectors $\displaystyle v_i$ by solving $\displaystyle Bv_i=\lambda_iv_i$ for $\displaystyle i \in\{1,2,3\}$. The matrix $\displaystyle P$ you're looking for will simply be a matrix which columns are the three eigenvectors.
• Aug 12th 2008, 06:49 AM
thanks for your help, however the problem there is so far ive found 2 different sets of eigenvalues, and i cannot find corresponding eigenvectors for either set. this is really important i get this right, my next year at university rests on ensuring answers to problems like these are correct.
• Aug 12th 2008, 07:03 AM
flyingsquirrel
Quote:

however the problem there is so far ive found 2 different sets of eigenvalues, and i cannot find corresponding eigenvectors for either set.

What did you find for $\displaystyle P(\lambda)$ ?
• Aug 12th 2008, 07:15 AM
do you mean the polynomial I found, - lambda cubed + 5lambda squared - 6lambda. (sorry i couldnt find the symbols)
however i have also read that for a triangular square matrix the eigenvalues are the diagonal values?
i then cannot find the eigenvectors?! ive been attempting this question for a number of days now, please help.
• Aug 12th 2008, 07:56 AM
flyingsquirrel
Quote:

do you mean the polynomial I found, - lambda cubed + 5lambda squared - 6lambda. (sorry i couldnt find the symbols)

That's it. Factoring the polynomial, we get
$\displaystyle -\lambda^3+5\lambda^2-6\lambda=-\lambda(\lambda^2-5\lambda+6)=-\lambda(\lambda-2)(\lambda-3)$
so the eigenvalues are 0, 2 and 3.
Quote:

however i have also read that for a triangular square matrix the eigenvalues are the diagonal values?
I didn't know that but that's right.
Quote:

i then cannot find the eigenvectors?! ive been attempting this question for a number of days now, please help.
Let's find the eigenvector $\displaystyle v=\begin{bmatrix}x\\y\\z\end{bmatrix}$ whose corresponding eigenvalue is 2.

We know that $\displaystyle Bv=2v \Longleftrightarrow \begin{bmatrix} 0& 1& 1\\ 0& 2& 2\\ 0& 0& 3 \end{bmatrix} \cdot \begin{bmatrix} x\\ y\\ z \end{bmatrix} =2\begin{bmatrix}x \\ y \\ z \end{bmatrix} \Longleftrightarrow \begin{bmatrix} y +z\\ 2y+2z\\ 3z \end{bmatrix} =\begin{bmatrix}2x \\ 2y \\ 2z \end{bmatrix}$ thus we have to solve $\displaystyle \begin{cases} y+z=2x&(1)\\ 2y+2z=2y&(2)\\ 3z=2z&(3)\\ \end{cases}$

From (3) we get that $\displaystyle z=0$, (2) is useless (0=0) and (1) tells us that $\displaystyle y=2x$ so $\displaystyle v=\begin{bmatrix}x\\2x\\0 \end{bmatrix}$ for $\displaystyle x\neq 0$. We can, for example, choose $\displaystyle x=1$ to get $\displaystyle v=\begin{bmatrix}1\\2\\0 \end{bmatrix}$.
• Aug 12th 2008, 12:02 PM
thank you so much for your help, however for eigenvalue 0 I am still struggling to find a non-zero eigenvector. I have the eigenvector for 3. When I then place the 3 eigenvectors into matrix form does it matter which column I place each vector in?!
• Aug 12th 2008, 12:24 PM
flyingsquirrel
Quote:

thank you so much for your help, however for eigenvalue 0 I am still struggling to find a non-zero eigenvector.

This time we have to solve
$\displaystyle \begin{cases} y+z=0&(1)\\ 2y+2z=0&(2)\\ 3z=0&(3)\\ \end{cases}$
From (3) we get that $\displaystyle z=0$, from (2) and (1) that $\displaystyle y=-z=0$. As there is no condition on $\displaystyle x$, the solutions are the vectors $\displaystyle \begin{bmatrix}x\\0\\0\end{bmatrix}$ for $\displaystyle x\neq 0$. If you choose, for example, $\displaystyle x=1$, you get a non-zero vector.

Quote:

I have the eigenvector for 3. When I then place the 3 eigenvectors into matrix form does it matter which column I place each vector in?!
No, it does not matter.
• Aug 13th 2008, 09:21 AM
thank you, however, now i have all 3 eigenvectors, no matter which way I place them into the matrix P I cannot find the diagonal vector D. i am always one number out.
• Aug 13th 2008, 10:44 AM
flyingsquirrel
Quote:

thank you, however, now i have all 3 eigenvectors, no matter which way I place them into the matrix P I cannot find the diagonal vector D. i am always one number out.

I don't understand how you can be "one number out", the diagonal entries of $\displaystyle D$ are the three eigenvalues of $\displaystyle B$.
• Aug 13th 2008, 11:34 AM
because BP does not equal PD

for BP the resultant matrix is
[0 2 3]
[0 4 6]
[0 0 0]

and for PD the resultant matrix is
[0 2 3]
[0 4 9]
[0 0 0]
• Aug 13th 2008, 11:53 AM
Opalg
Quote:

because BP does not equal PD

for BP the resultant matrix is
[0 2 3]
[0 4 6]
[0 0 0]

and for PD the resultant matrix is
[0 2 3]
[0 4 9]
[0 0 0]

Check your eigenvector for the eigenvalue 3. If you get it right, you should find that both BP and PD are $\displaystyle \begin{bmatrix}0&2&3\\0&4&6\\0&0&3\end{bmatrix}$.
• Aug 13th 2008, 12:01 PM