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Let $A$ be real symmetric $ntimes n$ matrix whose only eigenvalues are 0 and 1. Pick out the true statements.

How does the characteristics polynomial be $(lambda-1)^m lambda^{m-n}$ where $m le n$ here?Simplicity of eigenvalueProve that the eigenvalues of a real symmetric matrix are real.Let $A$ and $B$ be $n times n$ complex matrices. Pick out the true statements.Let $A$ be a $2 × 2$ matrix with real entries which is not a diagonal matrix and which satisfies $A^3 = I$. Pick out the true statements:(generalized) eigenvectorsDiagonalization & Algebraic MultiplicitiesPick out the correct choices for a real symmetric matrixCreative ways to show that a given matrix is diagonalizable?Orthogonality of the degenerate eigenvectors of a real symmetric matrixAlgebraic and geometric multiplicities of eigenvalues of a $3 times 3$ matrix

5

5

$begingroup$

Let $A$ be real symmetric $ntimes n$ matrix whose only eigenvalues are $0$ and $1$. Let the dimension of the null space of $A-I$ be $m$. Pick out the true statements.

1. The characteristic polynomial of $A$ is $(lambda-1)^m(lambda)^{m-n}$.




2. $A^k = A^{k+1}$




3. The rank of $A$ is $m$.

This is what I did: I found geometric multiplicity corresponding to eigenvalue value $0$ to be $n-m$(using symmetric matrix is diagonalizable ) while geometric multiplicity of eigenvalue value 1 is $m$(that is given) .so the characteristic polynomial of $A$ should be $(lambda-1)^m(lambda)^{n-m}$. While I am not sure about other two statements. Any kind of help is highly appreciated. Thanks.

linear-algebra matrices eigenvalues-eigenvectors

share|cite|improve this question

edited May 24 '13 at 6:03

learner

3,39032469

asked May 12 '12 at 13:07

srijansrijan

6,48063980

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you are right.
  $endgroup$
  – Ding Dong
  May 12 '12 at 13:38
  

  
  
  
  

add a comment | 

5

5

$begingroup$

Let $A$ be real symmetric $ntimes n$ matrix whose only eigenvalues are $0$ and $1$. Let the dimension of the null space of $A-I$ be $m$. Pick out the true statements.

1. The characteristic polynomial of $A$ is $(lambda-1)^m(lambda)^{m-n}$.




2. $A^k = A^{k+1}$




3. The rank of $A$ is $m$.

This is what I did: I found geometric multiplicity corresponding to eigenvalue value $0$ to be $n-m$(using symmetric matrix is diagonalizable ) while geometric multiplicity of eigenvalue value 1 is $m$(that is given) .so the characteristic polynomial of $A$ should be $(lambda-1)^m(lambda)^{n-m}$. While I am not sure about other two statements. Any kind of help is highly appreciated. Thanks.

linear-algebra matrices eigenvalues-eigenvectors

share|cite|improve this question

edited May 24 '13 at 6:03

learner

3,39032469

asked May 12 '12 at 13:07

srijansrijan

6,48063980

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you are right.
  $endgroup$
  – Ding Dong
  May 12 '12 at 13:38
  

  
  
  
  

add a comment | 

$begingroup$

Let $A$ be real symmetric $ntimes n$ matrix whose only eigenvalues are $0$ and $1$. Let the dimension of the null space of $A-I$ be $m$. Pick out the true statements.

1. The characteristic polynomial of $A$ is $(lambda-1)^m(lambda)^{m-n}$.




2. $A^k = A^{k+1}$




3. The rank of $A$ is $m$.

This is what I did: I found geometric multiplicity corresponding to eigenvalue value $0$ to be $n-m$(using symmetric matrix is diagonalizable ) while geometric multiplicity of eigenvalue value 1 is $m$(that is given) .so the characteristic polynomial of $A$ should be $(lambda-1)^m(lambda)^{n-m}$. While I am not sure about other two statements. Any kind of help is highly appreciated. Thanks.

linear-algebra matrices eigenvalues-eigenvectors

share|cite|improve this question

edited May 24 '13 at 6:03

learner

3,39032469

asked May 12 '12 at 13:07

srijansrijan

6,48063980

$endgroup$

share|cite|improve this question

edited May 24 '13 at 6:03

learner

3,39032469

asked May 12 '12 at 13:07

srijansrijan

6,48063980

share|cite|improve this question

edited May 24 '13 at 6:03

learner

3,39032469

asked May 12 '12 at 13:07

srijansrijan

6,48063980

edited May 24 '13 at 6:03

learner

3,39032469

edited May 24 '13 at 6:03

learner

3,39032469

asked May 12 '12 at 13:07

srijansrijan

6,48063980

asked May 12 '12 at 13:07

srijansrijan

6,48063980

2 Answers
2

active

oldest

votes

3

$begingroup$

As you noticed, since $A$ is a real symmetric matrix, it is diagonalizable, and the properties satisfied (or not) by $A$ will be satisfied also by $D$, where $D$ is a diagonal matrix which consists of the eigenvalues of $A$.
Indeed, $dimker(A-I)$ and characteristic polynomial are invariant up a change of basis. To see that, write $A=P^{-1}DP$, and $ker(A-I)=ker(D-I)$, as it can be seen by a double inclusion. For the characteristic polynomial, use determinant for example.

• The first is true if we switch $m$ and $n$ in the power of $lambda$, namely $p_A(lambda)=(lambda-1)^mlambda^{n-m}$, otherwise the degree would be $2m-n$, which is not $m$, except in the case $m=n$.


• The second is false if $k=0$ unless $A=I$, but true for $kgeq 1$.


• The third is also true. Indeed, we look at the rank of $A$ is the same as the rank of $D$, and the rank of a diagonal matrix is easy to determine.

share|cite|improve this answer

edited May 12 '12 at 14:27

answered May 12 '12 at 13:15

Davide GiraudoDavide Giraudo

128k17156268

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry sir i didn't get $dimker(A−I)$ and characteristic polynomial are invariant up a change of basis.
  $endgroup$
  – srijan
  May 12 '12 at 13:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And why third is true? Can you explain please?
  $endgroup$
  – srijan
  May 12 '12 at 13:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How second statement is true for all $kgeq 1$ ? Can you explain sir?
  $endgroup$
  – srijan
  May 12 '12 at 13:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  First show that it's enough to show it for $D$.
  $endgroup$
  – Davide Giraudo
  May 12 '12 at 13:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Dear sir i got all the points you made except for the first statement. My observation is characteristic polynomial of $A$ is $(lambda-1)^m(lambda)^{n-m}$. Does it make no difference ?
  $endgroup$
  – srijan
  May 12 '12 at 13:57
  

  
  
  
  

 | 
show 2 more comments

2

$begingroup$

To answer 2 and 3, consider diagonalizing $A$. And note that it is enough to show 2 for $k=1$.

share|cite|improve this answer

answered May 12 '12 at 13:17

Martin ArgeramiMartin Argerami

129k1184185

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Dear sir what i have observed is correct or not?
  $endgroup$
  – srijan
  May 12 '12 at 13:19
  

  
  
  
  

add a comment | 

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3

$begingroup$

As you noticed, since $A$ is a real symmetric matrix, it is diagonalizable, and the properties satisfied (or not) by $A$ will be satisfied also by $D$, where $D$ is a diagonal matrix which consists of the eigenvalues of $A$.
Indeed, $dimker(A-I)$ and characteristic polynomial are invariant up a change of basis. To see that, write $A=P^{-1}DP$, and $ker(A-I)=ker(D-I)$, as it can be seen by a double inclusion. For the characteristic polynomial, use determinant for example.

• The first is true if we switch $m$ and $n$ in the power of $lambda$, namely $p_A(lambda)=(lambda-1)^mlambda^{n-m}$, otherwise the degree would be $2m-n$, which is not $m$, except in the case $m=n$.


• The second is false if $k=0$ unless $A=I$, but true for $kgeq 1$.


• The third is also true. Indeed, we look at the rank of $A$ is the same as the rank of $D$, and the rank of a diagonal matrix is easy to determine.

share|cite|improve this answer

edited May 12 '12 at 14:27

answered May 12 '12 at 13:15

Davide GiraudoDavide Giraudo

128k17156268

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry sir i didn't get $dimker(A−I)$ and characteristic polynomial are invariant up a change of basis.
  $endgroup$
  – srijan
  May 12 '12 at 13:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And why third is true? Can you explain please?
  $endgroup$
  – srijan
  May 12 '12 at 13:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How second statement is true for all $kgeq 1$ ? Can you explain sir?
  $endgroup$
  – srijan
  May 12 '12 at 13:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  First show that it's enough to show it for $D$.
  $endgroup$
  – Davide Giraudo
  May 12 '12 at 13:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Dear sir i got all the points you made except for the first statement. My observation is characteristic polynomial of $A$ is $(lambda-1)^m(lambda)^{n-m}$. Does it make no difference ?
  $endgroup$
  – srijan
  May 12 '12 at 13:57
  

  
  
  
  

 | 
show 2 more comments

3

$begingroup$

As you noticed, since $A$ is a real symmetric matrix, it is diagonalizable, and the properties satisfied (or not) by $A$ will be satisfied also by $D$, where $D$ is a diagonal matrix which consists of the eigenvalues of $A$.
Indeed, $dimker(A-I)$ and characteristic polynomial are invariant up a change of basis. To see that, write $A=P^{-1}DP$, and $ker(A-I)=ker(D-I)$, as it can be seen by a double inclusion. For the characteristic polynomial, use determinant for example.

• The first is true if we switch $m$ and $n$ in the power of $lambda$, namely $p_A(lambda)=(lambda-1)^mlambda^{n-m}$, otherwise the degree would be $2m-n$, which is not $m$, except in the case $m=n$.


• The second is false if $k=0$ unless $A=I$, but true for $kgeq 1$.


• The third is also true. Indeed, we look at the rank of $A$ is the same as the rank of $D$, and the rank of a diagonal matrix is easy to determine.

share|cite|improve this answer

edited May 12 '12 at 14:27

answered May 12 '12 at 13:15

Davide GiraudoDavide Giraudo

128k17156268

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry sir i didn't get $dimker(A−I)$ and characteristic polynomial are invariant up a change of basis.
  $endgroup$
  – srijan
  May 12 '12 at 13:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And why third is true? Can you explain please?
  $endgroup$
  – srijan
  May 12 '12 at 13:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How second statement is true for all $kgeq 1$ ? Can you explain sir?
  $endgroup$
  – srijan
  May 12 '12 at 13:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  First show that it's enough to show it for $D$.
  $endgroup$
  – Davide Giraudo
  May 12 '12 at 13:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Dear sir i got all the points you made except for the first statement. My observation is characteristic polynomial of $A$ is $(lambda-1)^m(lambda)^{n-m}$. Does it make no difference ?
  $endgroup$
  – srijan
  May 12 '12 at 13:57
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

As you noticed, since $A$ is a real symmetric matrix, it is diagonalizable, and the properties satisfied (or not) by $A$ will be satisfied also by $D$, where $D$ is a diagonal matrix which consists of the eigenvalues of $A$.
Indeed, $dimker(A-I)$ and characteristic polynomial are invariant up a change of basis. To see that, write $A=P^{-1}DP$, and $ker(A-I)=ker(D-I)$, as it can be seen by a double inclusion. For the characteristic polynomial, use determinant for example.

• The first is true if we switch $m$ and $n$ in the power of $lambda$, namely $p_A(lambda)=(lambda-1)^mlambda^{n-m}$, otherwise the degree would be $2m-n$, which is not $m$, except in the case $m=n$.


• The second is false if $k=0$ unless $A=I$, but true for $kgeq 1$.


• The third is also true. Indeed, we look at the rank of $A$ is the same as the rank of $D$, and the rank of a diagonal matrix is easy to determine.

share|cite|improve this answer

edited May 12 '12 at 14:27

answered May 12 '12 at 13:15

Davide GiraudoDavide Giraudo

128k17156268

$endgroup$

share|cite|improve this answer

edited May 12 '12 at 14:27

answered May 12 '12 at 13:15

Davide GiraudoDavide Giraudo

128k17156268

answered May 12 '12 at 13:15

Davide GiraudoDavide Giraudo

128k17156268

answered May 12 '12 at 13:15

Davide GiraudoDavide Giraudo

128k17156268

2

$begingroup$

To answer 2 and 3, consider diagonalizing $A$. And note that it is enough to show 2 for $k=1$.

share|cite|improve this answer

answered May 12 '12 at 13:17

Martin ArgeramiMartin Argerami

129k1184185

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Dear sir what i have observed is correct or not?
  $endgroup$
  – srijan
  May 12 '12 at 13:19
  

  
  
  
  

add a comment | 

2

$begingroup$

To answer 2 and 3, consider diagonalizing $A$. And note that it is enough to show 2 for $k=1$.

share|cite|improve this answer

answered May 12 '12 at 13:17

Martin ArgeramiMartin Argerami

129k1184185

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Dear sir what i have observed is correct or not?
  $endgroup$
  – srijan
  May 12 '12 at 13:19
  

  
  
  
  

add a comment | 

$begingroup$

To answer 2 and 3, consider diagonalizing $A$. And note that it is enough to show 2 for $k=1$.

share|cite|improve this answer

answered May 12 '12 at 13:17

Martin ArgeramiMartin Argerami

129k1184185

$endgroup$

share|cite|improve this answer

answered May 12 '12 at 13:17

Martin ArgeramiMartin Argerami

129k1184185

answered May 12 '12 at 13:17

Martin ArgeramiMartin Argerami

129k1184185

answered May 12 '12 at 13:17

Martin ArgeramiMartin Argerami

129k1184185