Three problems of finding a base

[colt]

Homework Statement

Find The values of "Find the values of \upsilon for which the homogeneous
system (A - \upsilon I)X = 0 has non trivial solution e for these values of \upsilon,
find a base for the solution space from the system.
Which A being = \begin{pmatrix}<br /> 0 &amp; 0 &amp; 1 \\ 1 &amp; 0 &amp; -3 \\ 0 &amp; 1 &amp; 3 \end{pmatrix}

Homework Equations

The Attempt at a Solution

I already found the value of \upsilon which it is 1. My problem it is to find the base. I don't know how to do it in this case. Normally I pick up the matrix and scale it until I can find linearly independent vectors, but in this case, using the given matrix above I don't reach the correct solution. Should I scale this matrix?
\begin{pmatrix}<br /> 0 - \upsilon &amp; 0 &amp; 1 \\ 1 &amp; 0 - \upsilon &amp; -3 \\ 0 &amp; 1 &amp; 3 - \upsilon \end{pmatrix}. I scaled the normal matrix A

The correct answer it is: { (0,0,0,1),(1,0,1,0),(1,1,0,0) }

Homework Statement

Be M = { (3a + 4b - 4c, 2a -4b -6c, -2a -4b +2c) | a,b,c \epsilon R } a subspace of R³

a)Find a set of generators to M
b)Find a base for M

The Attempt at a Solution

Here another problem to find a base. I did part a) in which I found: { (3,2,-2),(4,-4,-4),(-4,6,2) }

Then I read the answer and saw { (3,2,-2),(1,-1,-1),(-2,-3,1) } Since the two last vectors that I found are multiple from these last two, I thought there would be no problem with my answer.

Then to solve part b) I picked up my vectors and scale them, trying to get the minimal amount of linearly independent vectors: \begin{pmatrix}<br /> 3 &amp; 4 &amp; -4 \\ 2 &amp; -4 &amp; -6 \\ -2 &amp; -4 &amp; 2 \end{pmatrix}. In the end the result that I found is wrong, but fairly close to the correct answer. So I think I did something wrong when constructing the matrix, but maybe it was just
arithmetic errors. So if someone can confirm if this is the right way to solve the problem, if it is I can show the scaling in details

Homework Statement

Find an orthonormal base for the solution set of the linear homogeneous system:
x - y - 2z +w = 0

The Attempt at a Solution

I try to construct three base vectors like this: x = y + 2z - w for y=1, z=0, w=0.
Then do the same for y=0, z=1, w=0. And finally for y=0, z=0, w=1.
After doing that I pick up three four dimensional vectors and check to see if they are orthogonal. I checked and saw that they aren't. Used orthogonal projection to construct an orthogonal vector to one of the original vectors and then normalized it...to see that it was wrong. So I ask: the procedure I did is right ? If it is I can put in details what I did. If not... what should I do?

Thanks for any input

 

[lurflurf]

1)Row reduce (A-v)

The correct answer it is not: (0,0,0,1),(1,0,1,0),(1,1,0,0)

 

[colt]

1)Row reduce (A-v)

The correct answer it is not: (0,0,0,1),(1,0,1,0),(1,1,0,0)

True I copied wrongly, it is (1,-2,1)

Anyway I am not being able to row reduce it. Here are my two attempts:

1°:
\begin{pmatrix}<br /> - \upsilon &amp; 0 &amp; 1 \\ 1 &amp; - \upsilon &amp; -3 \\ 0 &amp; 1 &amp; 3 - \upsilon \end{pmatrix}. L2 -> L2 + 3L1

\begin{pmatrix}<br /> - \upsilon &amp; 0 &amp; 1 \\ 1 - 3\upsilon &amp; - \upsilon &amp; 0 \\ 0 &amp; 1 &amp; 3 - \upsilon \end{pmatrix}

2°: \begin{pmatrix}<br /> - \upsilon &amp; 0 &amp; 1 \\ 1 &amp; - \upsilon &amp; -3 \\ 0 &amp; 1 &amp; 3 - \upsilon \end{pmatrix}. L2 -> L2 - \upsilon*L3

\begin{pmatrix}<br /> - \upsilon &amp; 0 &amp; 1 \\ 1 &amp; 0 &amp; -3 -3\upsilon -\upsilon^2 \\ 0 &amp; 1 &amp; 3 - \upsilon \end{pmatrix}

Then I don't know how to proceed in any of the cases

 

[Dick]

True I copied wrongly, it is (1,-2,1)

Anyway I am not being able to row reduce it. Here are my two attempts:

1°:
\begin{pmatrix}<br /> - \upsilon &amp; 0 &amp; 1 \\ 1 &amp; - \upsilon &amp; -3 \\ 0 &amp; 1 &amp; 3 - \upsilon \end{pmatrix}. L2 -> L2 + 3L1

\begin{pmatrix}<br /> - \upsilon &amp; 0 &amp; 1 \\ 1 - 3\upsilon &amp; - \upsilon &amp; 0 \\ 0 &amp; 1 &amp; 3 - \upsilon \end{pmatrix}

2°: \begin{pmatrix}<br /> - \upsilon &amp; 0 &amp; 1 \\ 1 &amp; - \upsilon &amp; -3 \\ 0 &amp; 1 &amp; 3 - \upsilon \end{pmatrix}. L2 -> L2 - \upsilon*L3

\begin{pmatrix}<br /> - \upsilon &amp; 0 &amp; 1 \\ 1 &amp; 0 &amp; -3 -3\upsilon -\upsilon^2 \\ 0 &amp; 1 &amp; 3 - \upsilon \end{pmatrix}

Then I don't know how to proceed in any of the cases

You said you had found the value of $\upsilon=1$ that will give you nontrivial solutions, presumably by solving the characteristic equation. So use it. Put $\upsilon=1$.