Similar Matrices have the same rank Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)prove the similar matrices have the same rankWhat subspace of 3 x 3 matrices is spanned by the invertible matrices? Rank 1 matrices? [GStrang P181, 3.5.29(a)(b)]Prove that matrices have equal rank.Can some one explain me a easy alternate proof of rank nullity theorem?Similar matrices of rank 1Two idempotent matrices are similar iff they have the same rankshow that a matrix a that is similar to an invertible matrix B is itself invertible. More generally, show that similar matrices have the same rank.Why “similar” matrices instead of “equivalent”?A and B are similar if and only if $textrank(A)=textrank(B)$?If $A$ is $ktimes l$ and $B$ is $ltimes k$, $lgeq k$ and full-rank, must $AB$ have full-rank?If real matrices $A,B$ have the same ch. polynomial and rank, and rk$(A-lambda I)= textrk(B-lambda I)$ for each eigenvalue, are they similar?
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Similar Matrices have the same rank
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)prove the similar matrices have the same rankWhat subspace of 3 x 3 matrices is spanned by the invertible matrices? Rank 1 matrices? [GStrang P181, 3.5.29(a)(b)]Prove that matrices have equal rank.Can some one explain me a easy alternate proof of rank nullity theorem?Similar matrices of rank 1Two idempotent matrices are similar iff they have the same rankshow that a matrix a that is similar to an invertible matrix B is itself invertible. More generally, show that similar matrices have the same rank.Why “similar” matrices instead of “equivalent”?A and B are similar if and only if $textrank(A)=textrank(B)$?If $A$ is $ktimes l$ and $B$ is $ltimes k$, $lgeq k$ and full-rank, must $AB$ have full-rank?If real matrices $A,B$ have the same ch. polynomial and rank, and rk$(A-lambda I)= textrk(B-lambda I)$ for each eigenvalue, are they similar?
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Prove that :- If $2$ matrices $A$ and $B$ are similar then they will have the same rank.
Proof is given here but I can't understand both answers which are related to image and kernel. I have seen all the video lectures of Prof. Gilbert Strang but I have not seen these things in those lectures. I only know that $A$ and $B$ are similar iff $A$ = $MBM^-1$ for some invertible square matrix $M$ but I can't proceed further. Is there any simple proof of it ?
Please help.
linear-algebra matrix-rank
asked Apr 2 at 8:33
ankitankit
1077
$endgroup$
add a comment |
$begingroup$
Prove that :- If $2$ matrices $A$ and $B$ are similar then they will have the same rank.
Proof is given here but I can't understand both answers which are related to image and kernel. I have seen all the video lectures of Prof. Gilbert Strang but I have not seen these things in those lectures. I only know that $A$ and $B$ are similar iff $A$ = $MBM^-1$ for some invertible square matrix $M$ but I can't proceed further. Is there any simple proof of it ?
Please help.
linear-algebra matrix-rank
asked Apr 2 at 8:33
ankitankit
1077
$endgroup$
1
$begingroup$
What definition of rank are you working with? Usually, rank is defined as the dimension of the image, so it's hard to talk about rank without talking about images...
$endgroup$
– 5xum
Apr 2 at 8:37
$begingroup$
Isn't the rank also the number of non zero eigen values? And similar matrices have the same spectrum
$endgroup$
– Fareed AF
Apr 2 at 8:39
1
$begingroup$
@FareedAF The rank is only the number of non zero eigen values if the matrices are square.
$endgroup$
– 5xum
Apr 2 at 8:47
$begingroup$
@5xum The OP is clearly talking about square matrices, otherwise the product $MBM^-1$ doesn’t make sense.
$endgroup$
– amd
Apr 2 at 17:28
add a comment |
$begingroup$
Prove that :- If $2$ matrices $A$ and $B$ are similar then they will have the same rank.
Proof is given here but I can't understand both answers which are related to image and kernel. I have seen all the video lectures of Prof. Gilbert Strang but I have not seen these things in those lectures. I only know that $A$ and $B$ are similar iff $A$ = $MBM^-1$ for some invertible square matrix $M$ but I can't proceed further. Is there any simple proof of it ?
Please help.
linear-algebra matrix-rank
asked Apr 2 at 8:33
ankitankit
1077
$endgroup$
Prove that :- If $2$ matrices $A$ and $B$ are similar then they will have the same rank.
Proof is given here but I can't understand both answers which are related to image and kernel. I have seen all the video lectures of Prof. Gilbert Strang but I have not seen these things in those lectures. I only know that $A$ and $B$ are similar iff $A$ = $MBM^-1$ for some invertible square matrix $M$ but I can't proceed further. Is there any simple proof of it ?
Please help.
linear-algebra matrix-rank
linear-algebra matrix-rank
asked Apr 2 at 8:33
ankitankit
1077
asked Apr 2 at 8:33
ankitankit
1077
asked Apr 2 at 8:33
ankitankit
1077
asked Apr 2 at 8:33
ankitankit
1077
asked Apr 2 at 8:33
ankitankit
1077
1077
1
$begingroup$
What definition of rank are you working with? Usually, rank is defined as the dimension of the image, so it's hard to talk about rank without talking about images...
$endgroup$
– 5xum
Apr 2 at 8:37
$begingroup$
Isn't the rank also the number of non zero eigen values? And similar matrices have the same spectrum
$endgroup$
– Fareed AF
Apr 2 at 8:39
1
$begingroup$
@FareedAF The rank is only the number of non zero eigen values if the matrices are square.
$endgroup$
– 5xum
Apr 2 at 8:47
$begingroup$
@5xum The OP is clearly talking about square matrices, otherwise the product $MBM^-1$ doesn’t make sense.
$endgroup$
– amd
Apr 2 at 17:28
add a comment |
1
$begingroup$
What definition of rank are you working with? Usually, rank is defined as the dimension of the image, so it's hard to talk about rank without talking about images...
$endgroup$
– 5xum
Apr 2 at 8:37
$begingroup$
Isn't the rank also the number of non zero eigen values? And similar matrices have the same spectrum
$endgroup$
– Fareed AF
Apr 2 at 8:39
1
$begingroup$
@FareedAF The rank is only the number of non zero eigen values if the matrices are square.
$endgroup$
– 5xum
Apr 2 at 8:47
$begingroup$
@5xum The OP is clearly talking about square matrices, otherwise the product $MBM^-1$ doesn’t make sense.
$endgroup$
– amd
Apr 2 at 17:28
1
1
$begingroup$
What definition of rank are you working with? Usually, rank is defined as the dimension of the image, so it's hard to talk about rank without talking about images...
$endgroup$
– 5xum
Apr 2 at 8:37
$begingroup$
What definition of rank are you working with? Usually, rank is defined as the dimension of the image, so it's hard to talk about rank without talking about images...
$endgroup$
– 5xum
Apr 2 at 8:37
$begingroup$
Isn't the rank also the number of non zero eigen values? And similar matrices have the same spectrum
$endgroup$
– Fareed AF
Apr 2 at 8:39
$begingroup$
Isn't the rank also the number of non zero eigen values? And similar matrices have the same spectrum
$endgroup$
– Fareed AF
Apr 2 at 8:39
1
1
$begingroup$
@FareedAF The rank is only the number of non zero eigen values if the matrices are square.
$endgroup$
– 5xum
Apr 2 at 8:47
$begingroup$
@FareedAF The rank is only the number of non zero eigen values if the matrices are square.
$endgroup$
– 5xum
Apr 2 at 8:47
$begingroup$
@5xum The OP is clearly talking about square matrices, otherwise the product $MBM^-1$ doesn’t make sense.
$endgroup$
– amd
Apr 2 at 17:28
$begingroup$
@5xum The OP is clearly talking about square matrices, otherwise the product $MBM^-1$ doesn’t make sense.
$endgroup$
– amd
Apr 2 at 17:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
My favourite proof goes along the lines of
$rk(B)geq rk(MBM^-1) = rk(A)$, as multiplying can only reduce rank (or keep it unchanged), never increase it. Now note that $B = M^-1AM$, so we similarily get $rk(A)geq rk(M^-1AM) = rk(B)$
answered Apr 2 at 8:38
ArthurArthur
123k7122211
$endgroup$
$begingroup$
Thank you! @Arthur
$endgroup$
– ankit
Apr 2 at 8:57
add a comment |
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$begingroup$
My favourite proof goes along the lines of
$rk(B)geq rk(MBM^-1) = rk(A)$, as multiplying can only reduce rank (or keep it unchanged), never increase it. Now note that $B = M^-1AM$, so we similarily get $rk(A)geq rk(M^-1AM) = rk(B)$
answered Apr 2 at 8:38
ArthurArthur
123k7122211
$endgroup$
$begingroup$
Thank you! @Arthur
$endgroup$
– ankit
Apr 2 at 8:57
add a comment |
$begingroup$
My favourite proof goes along the lines of
$rk(B)geq rk(MBM^-1) = rk(A)$, as multiplying can only reduce rank (or keep it unchanged), never increase it. Now note that $B = M^-1AM$, so we similarily get $rk(A)geq rk(M^-1AM) = rk(B)$
answered Apr 2 at 8:38
ArthurArthur
123k7122211
$endgroup$
$begingroup$
Thank you! @Arthur
$endgroup$
– ankit
Apr 2 at 8:57
add a comment |
$begingroup$
My favourite proof goes along the lines of
$rk(B)geq rk(MBM^-1) = rk(A)$, as multiplying can only reduce rank (or keep it unchanged), never increase it. Now note that $B = M^-1AM$, so we similarily get $rk(A)geq rk(M^-1AM) = rk(B)$
answered Apr 2 at 8:38
ArthurArthur
123k7122211
$endgroup$
My favourite proof goes along the lines of
$rk(B)geq rk(MBM^-1) = rk(A)$, as multiplying can only reduce rank (or keep it unchanged), never increase it. Now note that $B = M^-1AM$, so we similarily get $rk(A)geq rk(M^-1AM) = rk(B)$
answered Apr 2 at 8:38
ArthurArthur
123k7122211
answered Apr 2 at 8:38
ArthurArthur
123k7122211
answered Apr 2 at 8:38
ArthurArthur
123k7122211
answered Apr 2 at 8:38
ArthurArthur
123k7122211
123k7122211
$begingroup$
Thank you! @Arthur
$endgroup$
– ankit
Apr 2 at 8:57
add a comment |
$begingroup$
Thank you! @Arthur
$endgroup$
– ankit
Apr 2 at 8:57
$begingroup$
Thank you! @Arthur
$endgroup$
– ankit
Apr 2 at 8:57
$begingroup$
Thank you! @Arthur
$endgroup$
– ankit
Apr 2 at 8:57
add a comment |
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$begingroup$
What definition of rank are you working with? Usually, rank is defined as the dimension of the image, so it's hard to talk about rank without talking about images...
$endgroup$
– 5xum
Apr 2 at 8:37
$begingroup$
Isn't the rank also the number of non zero eigen values? And similar matrices have the same spectrum
$endgroup$
– Fareed AF
Apr 2 at 8:39
1
$begingroup$
@FareedAF The rank is only the number of non zero eigen values if the matrices are square.
$endgroup$
– 5xum
Apr 2 at 8:47
$begingroup$
@5xum The OP is clearly talking about square matrices, otherwise the product $MBM^-1$ doesn’t make sense.
$endgroup$
– amd
Apr 2 at 17:28