# Find matrix U

• May 11th 2008, 11:08 PM
deragon999
Find matrix U
Third part of the same question...the only part i cant seem to understand at all....

A matrix M=[3,1,0;1,2,1;0,1,3] where ; specifies a new row

The characteristic equation is f(x)=(3-x)(x-4)(x-1)

The eigenvalues for the characteristic equation are 3, 4 and 1?

I need to find a matrix, U so that (U^t)*M*U is a diagonal matrix
• May 11th 2008, 11:29 PM
Moo
Hello,

Quote:

Originally Posted by deragon999
Third part of the same question...the only part i cant seem to understand at all....

A matrix M=[3,1,0;1,2,1;0,1,3] where ; specifies a new row

The characteristic equation is f(x)=(3-x)(x-4)(x-1)

The eigenvalues for the characteristic equation are 3, 4 and 1?

I need to find a matrix, U so that (U^t)*M*U is a diagonal matrix

Find the eigenspaces/vectors for each eigenvalue (3, 4, 1). Do you know how to do it ?
Then, find the matrix transition between the canonical basis (assuming that M was written according to this basis) and the new basis, formed by the eigenvectors.
• May 12th 2008, 12:17 AM
Reckoner
Quote:

Originally Posted by deragon999
Third part of the same question...the only part i cant seem to understand at all....

A matrix M=[3,1,0;1,2,1;0,1,3] where ; specifies a new row

The characteristic equation is f(x)=(3-x)(x-4)(x-1)

The eigenvalues for the characteristic equation are 3, 4 and 1?

I need to find a matrix, U so that (U^t)*M*U is a diagonal matrix

We need $U^{\,T}MU$ to be diagonal.

You are looking for an orthogonal matrix $U$ (meaning that $U^{-1} = U^{\,T}$) that diagonalizes $M$. Find all linearly independent eigenvectors of $M$. Since $M$ is symmetric, any pair of eigenvectors that correspond to different eigenvalues will be orthogonal to each other. So, take your 3 eigenvectors, and normalize them to get an orthonormal set, which you can then place in the columns of $U$ to diagonalize $M$. Note that if your matrix had had an eigenvalue of multiplicity $k$ greater than 1, you would need to apply the Gram-Schmidt orthonormalization process on the $k$ linearly independent vectors corresponding to that eigenvalue in order to ensure that your set is orthonormal.

If you do the above, the columns (and rows) of $U$ will form an orthonormal set, so $U$ will be orthogonal, and $U^{-1}MU = U^{\,T}MU = D,\text{ where }D\text{ is diagonal}$.
• May 12th 2008, 03:54 AM
deragon999
From eigenvectors onwards
I have found the the three eigenvectors(i hope) which are:

v1=[1;0;-1] v2=[1;1;1] v3=[-1;2;-1]

I have used the following method on them: (<-- is the pivot row)

1 0 -1 <--
1 1 1
-1 2 -1
0 1 2 <--
0 2 -2
0 0 -6<--

Which, when the pivot rows are written as columns of a U, gives U as being:[1,0,0;0,1,0;-1,2,-6]

Which when applied to (U^t)*M*U this matrix, U, finds D to be:[6,-6,-18;-6,18,-42;-18,-42,108]
Which satisfies the question "Find a matrix(any matrix) U, so that (U^t)*M*U is diagonal"....correct?

Is this a valid method for working out a matrix, U?
• May 12th 2008, 01:13 PM
Moo
Quote:

Originally Posted by deragon999
I have found the the three eigenvectors(i hope) which are:

v1=[1;0;-1] v2=[1;1;1] v3=[-1;2;-1]

I have used the following method on them: (<-- is the pivot row)

1 0 -1 <--
1 1 1
-1 2 -1
0 1 2 <--
0 2 -2
0 0 -6<--

I find exactly the same eigenvectors (Clapping)

Now you have to normalize them...

$|v1|=\sqrt{1^2+0^2+(-1)^2}=\sqrt{2}$

-> v'1=v1/|v1|

$|v2|=\sqrt{3}$

$|v3|=\sqrt{6}$

Quote:

Which, when the pivot rows are written as columns of a U, gives U as being:[1,0,0;0,1,0;-1,2,-6]
Where does it come from ? o.O

Quote:

Which when applied to (U^t)*M*U this matrix, U, finds D to be:[6,-6,-18;-6,18,-42;-18,-42,108]
Which satisfies the question "Find a matrix(any matrix) U, so that (U^t)*M*U is diagonal"....correct?

Is this a valid method for working out a matrix, U?
Hm nope ~

You got U, formed by the normalised eigenvectors in columns (in the order 3-4-1) :

$U=\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{-1}{\sqrt{6}} \\
0 & \frac{1}{\sqrt{3}} & \frac{2}{\sqrt{6}} \\
\frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & \frac{-1}{\sqrt{6}} \end{pmatrix}$

$D=U^t M U= \begin{pmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

(in the same order as you put the eigenvectors in U)

(Whew) Hope it helps
• May 13th 2008, 12:25 AM
Dr Zoidburg
Is that correct for the eigenvectors?
I ask because I get (1, -2, 1) not (-1, 2, -1) for eigenvalue: 1

which of us is correct?
• May 13th 2008, 11:45 AM
Moo
Quote:

Originally Posted by Dr Zoidburg
Is that correct for the eigenvectors?
I ask because I get (1, -2, 1) not (-1, 2, -1) for eigenvalue: 1

which of us is correct?

The two of you :-)

The eigenvectors form basis, so you don't care wether it is -u or u.