Matrices (Generalised Inverse 2)

Consider the solution of the matrix system M x = r, where the matrix and right-hand side are given by

$M= \begin{pmatrix} 4& 4& 0\\4 & 5 & 2\\0 & 2& 4\end{pmatrix}$; $r = \begin{pmatrix} 2\\ 5\\ 6\end{pmatrix}$



The previous parts to this question was proving what the eigenvectors were (simple dot product) and writing down associated eigenvalues.

The question I'm stuck on:

We can write the vector x as a linear combination of the eigenvectors.

Writing $x = \sum\limits_{i} \beta_i V_i$; $r = \sum \limits_{i} \alpha_i V_i$, derive a formula for $\beta_i$ in terms of $\alpha_i$ and the eigenvalues $\lambda_i$

Edit: Is something up with latex? Wow...

Yes, something appears to be up in latex land...anyway, press on...

How far have you got? Have you derived the relation ?

Now you should notice that the $V_{i}$ are not linearly independent (as the matrix is singular). If $V_{3}$ is the eigenvector corresponding to the zero eigenvalue, what does that imply about $\beta_{3}$ ?

v3 is not the eigenvector that corresponds to the zero eigenvalue though?

I gave them an arbitrary ordering for convenience: v3 = (2,-2,1)

Ah, yeah that's my V1

I'm trying to solve to find the solution vector x but don't really understand my lecturers working (to a similar problem).
mx = r

OK, so I take v1 = (4,5,2); v2 = (-1,0,2) and v3 = (2,-2,1). Then by inspection, r = v1 + 2 v2. So you now have to solve

$\displaystyle \beta_{1} \lambda_{1} V_{1} + \beta_{2} \lambda_{2} V_{2}= V_{1} + 2 V_{2}$

remembering the comment about $\beta_{3}$.