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I am trying to find the upper triangular form of $B$ and an invertible matrix $C$ such that $B=C^{-1}AC$ where A is given by the following:

$A = \pmatrix{1&1\\ -1&3}$

The characteristic equation is $(x-2)^2$ with eigenvalue $2$ and eigenvector $\langle1, 1\rangle$.

  1. Finding an upper triangular matrix similar to $A$ means to find a vector $w$ such that $(A-2I)w = v$.
  2. In the above, is $v$ the eigenvector?

Could someone explain both of the above to me and why this is true?

Taking the above statements to be true we have:

$\pmatrix{1&-1 \\ 1&-1}\pmatrix{w_1 \\ w2} = \pmatrix{1 \\ 1}$

$\implies w_1 - w_2 = 1$ One solution is $\langle1, 0\rangle$. From here I do not know what to do next.

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I would recommend reading the Wikipedia entry on Jordan Normal Form. –  Michael Boratko Dec 18 '12 at 3:00
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1 Answer

up vote 4 down vote accepted

You want to find the Jordan Normal Form.

So, we are looking for an upper triangular matrix $J$ and an invertible matrix P s.t. $J=P^{-1}AP$ where: $A = \pmatrix{1&1\\ -1&3}$

The characteristic equation is found by solving $(A-\lambda I) = 0$, resulting in characteristic polynomial $(\lambda-2)^2 = 0$, giving us a repeated eigenvalue, $\lambda = 2$.

To find the first eigenvector, we use the eigenvalue $\lambda = 2$ and solve $(A-\lambda I)v1 = 0$, resulting in the eigenvector $v1 = \langle1, 1\rangle$.

Note: I am not sure if you have learned or understand algebraic and geometric multiplicity yet, so I'll stay away from that terminology, but it is very important! See Geometric and Algebraic Multiplicity.

Next, we need to find another eigenvector, but because we have a repeated eigenvalue, this is typically called a generalized eigenvector (if it exists from the Note above).

To find the generalized eigenvector, we can form $(A-\lambda I)v2 = v1$, so $\pmatrix{-1&1 \\ -1&1}\pmatrix{v21 \\ v22} = \pmatrix{1 \\ 1}$

$\implies v21 = v22-1$. Let $v22 = 0$, so $v21 = -1$, so the generalized eigenvector is $v2 = \langle-1, 0\rangle$.

Now, we can form $P = (v1 |\ v2) = \pmatrix{1&-1 \\ 1&0}$. That is, form the linear combination of $v1$ and $v2$.

Calculating $P^{-1} = \pmatrix{0&1 \\ -1&1}$, of course, this inverse is easy for a $2 x 2$ matrix since we have $P$.

Finally, we can calculate $J=P^{-1}AP = \pmatrix{2&1 \\ 0&2}$.

Do you notice anything special about the matrix $J$?

Let's verify these results using Wolfram Alpha.

Does this make sense and you follow?

Follow-Up Answer

How do you do this in general?

Not every matrix has enough linearly independent eigenvectors to be diagonalizable. However by using similarity transformations every square matrix can be transformed to the Jordan canonical form, which is almost diagonal.

See these wonderful notes (particularly at 7.2.1 for a sample 4x4, which shows all of the possibilities for a 4x4, but read it all and go through all of the examples with algebraic and geometric multiplicity), Jordan Canonical Form.

The general condition is the presence of nontrivial Jordon Blocks, see Jordan Normal Form- look at the 4x4 example and work it.

Also look at this interesting post Number of Jordan canonical forms for an nxn matrix.

There are more examples for you to review here.

Additionally, you might find these helpful.

Diagonalizable Matrices: How to determine?

When is the geometric multiplicity of an eigenvalue smaller than its algebraic multiplicity?

Can an eigenvalue (of an $n$ by $n$ matrix A) with algebraic multiplicity $n$ have an eigenspace with fewer than $n$ dimensions?

See Jordan Form Section

Regards - Amzoti

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What do you do in general $(n\times n)$ matrix? Suppose I have a $3\times 3$ matrix. Instead of only computing $(A-\lambda I)v_2=v_1$, what else would you compute? –  CodeKingPlusPlus Dec 19 '12 at 5:11
    
Please see my Follow-Up answer given that it was too large for comment. Regards. –  Amzoti Dec 19 '12 at 7:36
    
Great job! Such a nice write up with links for additional support! $+1\quad\large\checkmark$ –  amWhy May 12 '13 at 0:56
    
Ah! I was looking for that answer the other day to reuse it! I have such a hard time searching for stuff on this site, wish there was a way to improve that - including to look for duplicates! Thx –  Amzoti May 12 '13 at 1:00
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