Thread: 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

Hello,

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.

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 to be diagonal.

You are looking for an orthogonal matrix (meaning that ) that diagonalizes . Find all linearly independent eigenvectors of . Since 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 to diagonalize . Note that if your matrix had had an eigenvalue of multiplicity greater than 1, you would need to apply the Gram-Schmidt orthonormalization process on the 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 will form an orthonormal set, so will be orthogonal, and .

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?

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

Now you have to normalize them...



-> v'1=v1/|v1|

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

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) :






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


Hope it helps

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?

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.