Question

Can soomeone help me calculate the bases of matrix
2 0 0 -6
0 2 0 -3
-3 -3 -1 -9
0 0 0 -1

with lamda = -1 & -2

Answer 1

Do you mean find the eigenvectors?

Answer 2

Well i need to diagonalize the matrix, I have the answer, but i don't know how to get it

Answer 3

Did you already find the eigenvalues to be -1 and -2?

Answer 4

The answers for lamda =-1
v1=
2
1
0
1

Answer 5

and v2=
0
0
1
0

Answer 6

no its positive 2

Answer 7

And for lamda = 2
1
-1
0
0
and
0
1
-1
0

Answer 8

The problem is i have no idea how to get the answer when i have lines with 0's

Answer 9

So then your D=(P^-1)A P

Where P is the matrix with those vectors as columns.
Your D

Answer 10

i

Answer 11

your D will be the corresponding eigenvalues on the diagonal.
(corresponding to which vector you put where in P)

Answer 12

righ like
-1 000
0-100
0020
0002

Answer 13

my p is equal to
2 0 1 0
1 0 -1 1
0 1 0 -1
1 0 0 0

Answer 14

the 2 last colonms, after i apply the lamda = 2 I get a reduced matrix of
1 1 1 0
0 0 0 1
0000
0000

Answer 15

The answer is

1
-1
0
0

Answer 16

and
0
1
-1
0

Answer 17

I have no idea how to get those 2 last vectors

Answer 18

I'm confused as to where you are having difficulty.
Were you able to find a basis for the eigenvalue of -1?

Answer 19

The answsers are there, I just don't know how to find the base of 2

Answer 20

The method used to get the 2 bases of the matrix

Answer 21

yea by lamda i meant eigenvectors sorry

Answer 22

ok lets work on that

Answer 23

the bases of -1 were
2
1
0
1
and
0
0
1
0

Answer 24

So for lamda =2 our matrix is:
0 0 0 6
0 0 0 3
3 3 3 9
0 0 0 3
Right?

Answer 25

Thats right

Answer 26

I reduce them to
1 1 1 0
0 0 0 1
0 0 0 0
0 0 0 0

Answer 27

From there, When theres more than one base to find i'm not sure what to do

Answer 28

Oh ok so we have leading 1s in which columns?

Answer 29

first and second?

Answer 30

No those are the rows with leading columns.

Answer 31

I mean those are the rows with leading ones.

Answer 32

Columns 1 and 4 have leading ones.

Answer 33

So far i follow

Answer 34

So my base is constructed by x1 and x4 ?

Answer 35

This tells us that
x4=0
x3=? so lets call it "t"
x2=? so lets call it "s"
x1+x2+x3=0

Answer 36

Agreed

Answer 37

So
x1=-s -t
x2= s
x3= t
x4 = 0

Answer 38

xn=(-1,1,0,0)s+(-1,0,1,0)t

Answer 39

Thats where I don't know what to do

Answer 40

\[\left[\begin{matrix}x1 \\ x2\\ x3\\x4\end{matrix}\right]=\left[\begin{matrix}-s -t \\ s\\ t\\0\end{matrix}\right]=\left[\begin{matrix}-s -t \\ s+0\\0+ t\\0+0\end{matrix}\right]=\left[\begin{matrix}-1 \\ 1\\ 0\\0\end{matrix}\right]s+\left[\begin{matrix}-1 \\ 0\\ 1\\0\end{matrix}\right]t\]

Answer 41

The answer I was given is
1
-1
0
0
and
0
1
-1
0

Answer 42

Well our first element of the basis matches because that is just (-1) times what I put.
I don't know about that 2nd one. Where did you get the answer?

Answer 43

In a lot of different problems , I get to the step you wrote, but that isn't the answer according to my solutions

Answer 44

Its the sollutions from past exams .. I could send you the pdfs but Its in french

Answer 45

Its the first question

Answer 46

Both vectors are correct.
That is just a different basis for the same space.

Answer 47

Do you know how the vectors were obtained?
Because if I do my verifications i don't gt the same answers

Answer 48

If you subtract our v2 by our v1 you will get their v2.

Answer 49

it is in french!>>>???

Answer 50

Yes

Answer 51

you know how to speak french?

Answer 52

I do

Answer 53

download an english linear algebra

Answer 54

you can check your bases elements by multiplying them by the matrix we row reduced to find them.
0 0 0 6
0 0 0 3
3 3 3 9
0 0 0 3

Answer 55

But are those vectors linearly independant ?

Answer 56

the product should be the zero vector

Answer 57

They are always linearly independent if they are the solutions of a homogeneous system.

Answer 58

Also the eigenvector bases of distinct eigenvalues are orthoganal to each other, but the elements within each bases may not be (but they will be lin indep).

Answer 59

I'll verify them right away

Answer 60

I verified the ones for lamda = 2 already.
We are correct.

Answer 61

Your textbook might benefit from a disclaimer
"answer not unique".

Answer 62

I verified them , and they check out, but in a lot of cases, the answers are written in the the form that i wrote up, and not the ones that you showed me. Even though the answers check out, to be sure to get the aforementioned vectors, we do v2-v1 and what else for the v1? change the signs?

Answer 63

You don't need to get the same as the textbook.
I don't understand your latest question?

Answer 64

To get v2, the textbook answer, we did v2-v1
to get the v1 answer, we do v1=-v1?

Answer 65

There is nothing special about the textbook answer.
There are infinite bases to describe any vector space (all the linear combinations of the elements in them).
If your bases satisfies the multiplication by the matrix = zero vector that is the quickest way to check.
Otherwise you are going to have to find a linear combination of your bases that equals each texbook element. That is inefficient.

Answer 66

But the thing is that those pproblems were corrections from previous exams , and orur teacher wrote it that way, ... im guessing as long as \[P ^{-1}AP\] checks out theresholdnt be any problem right ?

Answer 67

Thanks!

Answer 68

Yes if that = a diagonal matrix then you are golden.