How to find multiplicity of eigenvalues?
$begingroup$
So i have this matrix:
enter image description here
Sorry for the formatting, I'm new here and any help would be great.
We compute the characteristic polynomial
$p(lambda)= {rm det}(A − lambda I d)= (lambda + 2)^2 (1 − λ)$
So from here I know that one of the root is 1 and the other one is -2. However, it states on the answer sheet that the multiplicity of -2 is 2 ? How come ? I don't see two -2 roots in the equation .
Somebody please help?
matrices matrix-equations matrix-calculus diagonalization
edited Nov 30 '18 at 19:08
David G. Stork
10.2k21332
asked Nov 30 '18 at 18:57
BM97BM97
758
$endgroup$
add a comment |
$begingroup$
So i have this matrix:
enter image description here
Sorry for the formatting, I'm new here and any help would be great.
We compute the characteristic polynomial
$p(lambda)= {rm det}(A − lambda I d)= (lambda + 2)^2 (1 − λ)$
So from here I know that one of the root is 1 and the other one is -2. However, it states on the answer sheet that the multiplicity of -2 is 2 ? How come ? I don't see two -2 roots in the equation .
Somebody please help?
matrices matrix-equations matrix-calculus diagonalization
edited Nov 30 '18 at 19:08
David G. Stork
10.2k21332
asked Nov 30 '18 at 18:57
BM97BM97
758
$endgroup$
add a comment |
$begingroup$
So i have this matrix:
enter image description here
Sorry for the formatting, I'm new here and any help would be great.
We compute the characteristic polynomial
$p(lambda)= {rm det}(A − lambda I d)= (lambda + 2)^2 (1 − λ)$
So from here I know that one of the root is 1 and the other one is -2. However, it states on the answer sheet that the multiplicity of -2 is 2 ? How come ? I don't see two -2 roots in the equation .
Somebody please help?
matrices matrix-equations matrix-calculus diagonalization
edited Nov 30 '18 at 19:08
David G. Stork
10.2k21332
asked Nov 30 '18 at 18:57
BM97BM97
758
$endgroup$
So i have this matrix:
enter image description here
Sorry for the formatting, I'm new here and any help would be great.
We compute the characteristic polynomial
$p(lambda)= {rm det}(A − lambda I d)= (lambda + 2)^2 (1 − λ)$
So from here I know that one of the root is 1 and the other one is -2. However, it states on the answer sheet that the multiplicity of -2 is 2 ? How come ? I don't see two -2 roots in the equation .
Somebody please help?
matrices matrix-equations matrix-calculus diagonalization
matrices matrix-equations matrix-calculus diagonalization
edited Nov 30 '18 at 19:08
David G. Stork
10.2k21332
asked Nov 30 '18 at 18:57
BM97BM97
758
edited Nov 30 '18 at 19:08
David G. Stork
10.2k21332
asked Nov 30 '18 at 18:57
BM97BM97
758
edited Nov 30 '18 at 19:08
David G. Stork
10.2k21332
edited Nov 30 '18 at 19:08
David G. Stork
10.2k21332
edited Nov 30 '18 at 19:08
David G. Stork
10.2k21332
10.2k21332
asked Nov 30 '18 at 18:57
BM97BM97
758
asked Nov 30 '18 at 18:57
BM97BM97
758
asked Nov 30 '18 at 18:57
BM97BM97
758
758
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Think of your characteristic equation as:
$$p(lambda) = (lambda + 2)(lambda + 2)(1 - lambda)$$
to see the multiplicity of eigenvalues better.
answered Nov 30 '18 at 19:01
David G. StorkDavid G. Stork
10.2k21332
$endgroup$
$begingroup$
So i would need to look at (λ + 2)² as it if were two distinct equations?
$endgroup$
– BM97
Nov 30 '18 at 19:02
$begingroup$
No... having two (equal) solutions.
$endgroup$
– David G. Stork
Nov 30 '18 at 19:06
$begingroup$
I don't understand ; i meant i need to see (λ + 2)² as two different brackets as you said
$endgroup$
– BM97
Nov 30 '18 at 19:08
$begingroup$
Are you unfamiliar with the fact that $(lambda + 2)^2 = (lambda +2)(lambda + 2)$?????
$endgroup$
– David G. Stork
Nov 30 '18 at 19:09
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Think of your characteristic equation as:
$$p(lambda) = (lambda + 2)(lambda + 2)(1 - lambda)$$
to see the multiplicity of eigenvalues better.
answered Nov 30 '18 at 19:01
David G. StorkDavid G. Stork
10.2k21332
$endgroup$
$begingroup$
So i would need to look at (λ + 2)² as it if were two distinct equations?
$endgroup$
– BM97
Nov 30 '18 at 19:02
$begingroup$
No... having two (equal) solutions.
$endgroup$
– David G. Stork
Nov 30 '18 at 19:06
$begingroup$
I don't understand ; i meant i need to see (λ + 2)² as two different brackets as you said
$endgroup$
– BM97
Nov 30 '18 at 19:08
$begingroup$
Are you unfamiliar with the fact that $(lambda + 2)^2 = (lambda +2)(lambda + 2)$?????
$endgroup$
– David G. Stork
Nov 30 '18 at 19:09
add a comment |
$begingroup$
Think of your characteristic equation as:
$$p(lambda) = (lambda + 2)(lambda + 2)(1 - lambda)$$
to see the multiplicity of eigenvalues better.
answered Nov 30 '18 at 19:01
David G. StorkDavid G. Stork
10.2k21332
$endgroup$
$begingroup$
So i would need to look at (λ + 2)² as it if were two distinct equations?
$endgroup$
– BM97
Nov 30 '18 at 19:02
$begingroup$
No... having two (equal) solutions.
$endgroup$
– David G. Stork
Nov 30 '18 at 19:06
$begingroup$
I don't understand ; i meant i need to see (λ + 2)² as two different brackets as you said
$endgroup$
– BM97
Nov 30 '18 at 19:08
$begingroup$
Are you unfamiliar with the fact that $(lambda + 2)^2 = (lambda +2)(lambda + 2)$?????
$endgroup$
– David G. Stork
Nov 30 '18 at 19:09
add a comment |
$begingroup$
Think of your characteristic equation as:
$$p(lambda) = (lambda + 2)(lambda + 2)(1 - lambda)$$
to see the multiplicity of eigenvalues better.
answered Nov 30 '18 at 19:01
David G. StorkDavid G. Stork
10.2k21332
$endgroup$
Think of your characteristic equation as:
$$p(lambda) = (lambda + 2)(lambda + 2)(1 - lambda)$$
to see the multiplicity of eigenvalues better.
answered Nov 30 '18 at 19:01
David G. StorkDavid G. Stork
10.2k21332
answered Nov 30 '18 at 19:01
David G. StorkDavid G. Stork
10.2k21332
answered Nov 30 '18 at 19:01
David G. StorkDavid G. Stork
10.2k21332
answered Nov 30 '18 at 19:01
David G. StorkDavid G. Stork
10.2k21332
10.2k21332
$begingroup$
So i would need to look at (λ + 2)² as it if were two distinct equations?
$endgroup$
– BM97
Nov 30 '18 at 19:02
$begingroup$
No... having two (equal) solutions.
$endgroup$
– David G. Stork
Nov 30 '18 at 19:06
$begingroup$
I don't understand ; i meant i need to see (λ + 2)² as two different brackets as you said
$endgroup$
– BM97
Nov 30 '18 at 19:08
$begingroup$
Are you unfamiliar with the fact that $(lambda + 2)^2 = (lambda +2)(lambda + 2)$?????
$endgroup$
– David G. Stork
Nov 30 '18 at 19:09
add a comment |
$begingroup$
So i would need to look at (λ + 2)² as it if were two distinct equations?
$endgroup$
– BM97
Nov 30 '18 at 19:02
$begingroup$
No... having two (equal) solutions.
$endgroup$
– David G. Stork
Nov 30 '18 at 19:06
$begingroup$
I don't understand ; i meant i need to see (λ + 2)² as two different brackets as you said
$endgroup$
– BM97
Nov 30 '18 at 19:08
$begingroup$
Are you unfamiliar with the fact that $(lambda + 2)^2 = (lambda +2)(lambda + 2)$?????
$endgroup$
– David G. Stork
Nov 30 '18 at 19:09
$begingroup$
So i would need to look at (λ + 2)² as it if were two distinct equations?
$endgroup$
– BM97
Nov 30 '18 at 19:02
$begingroup$
So i would need to look at (λ + 2)² as it if were two distinct equations?
$endgroup$
– BM97
Nov 30 '18 at 19:02
$begingroup$
No... having two (equal) solutions.
$endgroup$
– David G. Stork
Nov 30 '18 at 19:06
$begingroup$
No... having two (equal) solutions.
$endgroup$
– David G. Stork
Nov 30 '18 at 19:06
$begingroup$
I don't understand ; i meant i need to see (λ + 2)² as two different brackets as you said
$endgroup$
– BM97
Nov 30 '18 at 19:08
$begingroup$
I don't understand ; i meant i need to see (λ + 2)² as two different brackets as you said
$endgroup$
– BM97
Nov 30 '18 at 19:08
$begingroup$
Are you unfamiliar with the fact that $(lambda + 2)^2 = (lambda +2)(lambda + 2)$?????
$endgroup$
– David G. Stork
Nov 30 '18 at 19:09
$begingroup$
Are you unfamiliar with the fact that $(lambda + 2)^2 = (lambda +2)(lambda + 2)$?????
$endgroup$
– David G. Stork
Nov 30 '18 at 19:09
add a comment |
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