# Thread: diagonalising a 3x3 matrix

1. ## 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

2. Hello,
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 $B$ you first need to find its eigenvalues : the roots of the polynomial $P(\lambda)=\det(B-\lambda I_3)$. Once you know the three eigenvalues $\lambda_1,\lambda_2$ and $\lambda_3$, you can find the eigenvectors $v_i$ by solving $Bv_i=\lambda_iv_i$ for $i \in\{1,2,3\}$. The matrix $P$ you're looking for will simply be a matrix which columns are the three eigenvectors.

3. 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.

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 $P(\lambda)$ ?

5. 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.

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

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

From (3) we get that $z=0$, (2) is useless (0=0) and (1) tells us that $y=2x$ so $v=\begin{bmatrix}x\\2x\\0 \end{bmatrix}$ for $x\neq 0$. We can, for example, choose $x=1$ to get $v=\begin{bmatrix}1\\2\\0 \end{bmatrix}$.

7. 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?!

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

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.

9. 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.

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 $D$ are the three eigenvalues of $B$.

11. 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]

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 $\begin{bmatrix}0&2&3\\0&4&6\\0&0&3\end{bmatrix}$.

13. i found my mistake, thank you very much to both of you for all your help!!
the people on this forum are brilliant!! thanks again!

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# diagonalizable matrix example 3x3

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