Is the matrix Diagonalizable

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August 26th 2009, 12:39 PM
sofwhere

Is the matrix Diagonalizable

Hi I have a problem in Diagonalizability concept.

A= [7 -4 0
5 -5 0
6 -6 3]

A is 3x3 matrix.

I got eigen values as 3, 3, -1
Need to find eigen vectors and matrix D, where [D= Q^(-1)AQ]
D being the diagonal matrix.

If there is any easy methodology plz suggest

Need help from eigen vectors onwards.

Thanks in advance.
August 26th 2009, 12:56 PM
HallsofIvy

Quote:

Originally Posted by sofwhere

Hi I have a problem in Diagonalizability concept.

A= [7 -4 0
5 -5 0
6 -6 0]

A is 3x3 matrix.

I got eigen values as 3, [1+ 2 sqrroot(15)], 1- 2 sqrroot(15)]

Well, you got wrong then! Without doing any calculations, your matrix has a column of 0's so its determinant is 0 which means that it must have 0 as an eigenvalue.
The eigenvalue equation is $\left|\begin{array}{ccc}7-\lambda & -4 & 0 \\ 5 & -5-\lambda & 0 \\ 6 & -6 & -\lambda\end{array}\right|= 0$ . Expanding on the last column, that is $-\lambda \left|\begin{array}{cc} 7- \lambda & -4 \\ 5 & -5-\lambda\end{array}\right|$ $= (-\lambda)(\lambda^2- 2\lambda- 35- 20)$ $=(-\lambda)(\lambda^2- 2\lambda- 55)= 0$ . Thus, the eigenvalues are 0, $1+ \sqrt{56}= 1+ 2\sqrt{14}$ , and $1- \sqrt{56}= 1- 2\sqrt{14}$ .

Quote:

Need to find eigen vectors and matrix D, where [D= Q^(-1)AQ]
D being the diagonal matrix.

If there is any easy methodology plz suggest

Need help from eigen vectors onwards.

Thanks in advance.
August 26th 2009, 01:02 PM
Defunkt

Let $A = \left[ \begin{array}{ c c c } 7 & -4 & 0 \\ 5 & -5 & 0 \\ 6 & -6 & 0 \\ \end{array} \right]$

We can see that $rank(A) = 2$ , thus 0 is an eigenvalue. If you solve $det(A - \lambda I)=0$ , you will get that -3, 5 are the remaining eigenvalues.

Now, the diagonal matrix $D$ which satisfies $Q^{-1}AQ = D$ simply has $A$ 's eigenvalues as its diagonal entries. Do you understand why?

To find the eigenvectors for 0, -3, 5 -- you need to solve $(A - \lambda I)x = 0$ for $\lambda = 0, -3, 5.$

@HallsofIvy: $= (-\lambda)(\lambda^2- 2\lambda- 35- 20)$ should be = $(-\lambda)(\lambda^2- 2\lambda - 35 + 20)$ ; you're also missing an $=$ sign!
August 26th 2009, 01:13 PM
sofwhere

Is the matrix Diagonalizable

Sorry to say the element in the matrix is not zero. Its 3.
I also updated in my first post.

Also 2nd row first element is 8.
(Bow)
August 26th 2009, 01:14 PM
sofwhere

A= [7 -4 0
8 -5 0
6 -6 3]
August 26th 2009, 01:21 PM
Defunkt

Well, then the new eigenvalues are $3,3,-1$ , so $D$ will have them as its diagonal entries. To find the eigenvectors now, solve $(A-\lambda I)x = 0$ for $\lambda = 3,1$

(this is for $A = \left[ \begin{array}{ c c c } 7 & -4 & 0 \\ 8 & -5 & 0 \\ 6 & -6 & 3 \\ \end{array} \right]$ )
August 26th 2009, 01:58 PM
sofwhere

lambda(denote it as t) = 3,-1 ,3
Then i went for finding eigen vectors

B1= [A-(t1)I]

Then for t2=-1
B2= [ 8 -4 0
8 -4 0
6 -6 4 ]
the coefficients X= [x1, x2, x3]

[A-(t2)I]X= 0

need to find values of x1,x2,x3
August 26th 2009, 06:04 PM
HallsofIvy

Quote:

Originally Posted by sofwhere

lambda(denote it as t) = 3,-1 ,3
Then i went for finding eigen vectors

B1= [A-(t1)I]

Then for t2=-1
B2= [ 8 -4 0
8 -4 0
6 -6 4 ]
the coefficients X= [x1, x2, x3]

[A-(t2)I]X= 0

need to find values of x1,x2,x3

So, for eigenvalue -1, you want to solve 8x- 4y= 0, 8x- 4y= 0, and 6x- 6y+ 4z= 0. Those first two equations are obviously the same so they just say that y= 2x. Putting that into the last equation, 6x- 6(2x)+ 4z= 0 or -6x+ 4z= 0 or z= (6/4)x= (3/2)x. An eigenvector corresponding to eigenvalue -1 is x<1, 2, 3/2> and, if we take x= 2, that is <2, 4, 3>.

For eigenvalue 3, We have $A- 3I= \begin{bmatrix}7-3 & -4 & 0 \\ 8 & -5- 3 & 0 \\ 6 & -6 & 3-3\end{bmatrix}= \begin{bmatrix}4 & -4 & 0 \\ 8 & -8 & 0 \\ 6 & - 6 & 0\end{bmatrix}$ so you want to solve 4x- 4y= 0, 8x- 8y= 0, and 6x- 6y= 0. All of those, of course, reduce to y= x so any vector of the form <x, x, 0>= x<1, 1, 0> is an eigenvector corresponding to eigen value 3. But there was no "z" in those equations so z can be anything. <0, 0, z>= z< 0, 0, 1> is also an eigenvector corresponding to eigenvalue 3.

The fact that this matrix has 3 independent eigenvectors is what tells us it can be diagonalized. Let B be the matrix having those eigenvectors as columns and you should get $B^{-1}AB= \begin{bmatrix}-1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{bmatrix}$ (the order of numbers on the diagonal will depend on the order of the eigenvectors as columns).
August 27th 2009, 04:35 AM
sofwhere

I took D = Q^(-1)AQ

Where Q is matrix of eigen vectors (Its the B u referred)
I took

Q= [ 1 1 0
2 1 0
3/2 0 1]

and so;ved the equation by finding Q^(-1) and then Q^(-1)AQ

I got D= [-1 0 0
0 3 0
9/4 0 3]
Is this correct?

Or should I take the eigen vector for lambda= -1 as ( 2,4,3) as u mentioned and redo.
August 27th 2009, 05:49 AM
Defunkt

Quote:

Originally Posted by sofwhere

I took D = Q^(-1)AQ

Where Q is matrix of eigen vectors (Its the B u referred)
I took

Q= [ 1 1 0
2 1 0
3/2 0 1]

and so;ved the equation by finding Q^(-1) and then Q^(-1)AQ

I got D= [-1 0 0
0 3 0
9/4 0 3]
Is this correct?

Or should I take the eigen vector for lambda= -1 as ( 2,4,3) as u mentioned and redo.

What you should have gotten is this: $D = \begin{bmatrix}-1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{bmatrix}$ . The eigenvectors you take are irrelevant to your solution, as they all should produce the same one.

You can also find $D$ without directly calculating $Q^{-1}AQ$ -- it is a direct result of the definition of eigenvectors and eigenvalues.
August 27th 2009, 05:50 AM
courteous

Quote:

Originally Posted by sofwhere

lambda(denote it as t) = 3,-1 ,3
Then i went for finding eigen vectors

B1= [A-(t1)I]

Then for t2=-1
B2= [ 8 -4 0
8 -4 0
6 -6 4 ]
the coefficients X= [x1, x2, x3]

[A-(t2)I]X= 0

need to find values of x1,x2,x3

sofwhere, isn't the last (or the first(Happy)) eigenvalue -3 (instead of +3)?
August 27th 2009, 06:00 AM
sofwhere

Finally det(T-tI)= (3-t)(t+1)(t-3) = O

therefore t1=3, t2=-1, t3=3
These r the three eigen values and eigen vectors are

v1= t( 1 1 0)

v2= t( 1 2 3/2 )

v3 = t( 0 0 1)
where tbelongs to R

This is how i started.
August 28th 2009, 08:17 AM
sofwhere

Quote:

Originally Posted by courteous

sofwhere, isn't the last (or the first(Happy)) eigenvalue -3 (instead of +3)?

Thanks guyz finally got it.