Mixing
For $n$ by $n$ matrix A, prove if dimension of row space > $n/2$, then $A^2 ne 0$
$begingroup$
I'm trying to prepare for an exam and came across the following question:
Given $n times n$ matrix A, let U represent row space and W
represent column space.
A) Prove: $W subseteq$ $U^{perp} $ if and only if $A^2=0$
B) Prove that if $dim(U)>n/2$, then $A^2 ne 0$.
I was able to do part A pretty easily as follows:
$Rightarrow$: Let rows of A be {$ vec{r}_1, vec{r}_2 ,...,vec{r}_n $} and columns be {$ vec{c}_1, vec{c}_2 ,...,vec{c}_n $}. Assume $W subseteq$ $U^{perp} $. Entry $a_{ij}$ (in row i, column j of matrix $A^2=AA$) equals the dot product of $vec{r}_i cdot vec{c}_j $, and by assumption $vec{c}_j$ is orthogonal to $ vec{r}_1, vec{r}_2 ,...,vec{r}_n $, so it follows that $a_{ij}=0$ for all i,j. Thus $A^2=0$
$Leftarrow$: Assume $A^2=0$. We can represent the columns of $A^2$ as
$$ pmatrix{vec{r}_1 \ vec{r}_2 \ vdots \ vec{r}_n} cdot vec{c}_1 , pmatrix{vec{r}_1 \ vec{r}_2 \ vdots \ vec{r}_n} cdot vec{c}_2 , ... , pmatrix{vec{r}_1 \ vec{r}_2 \ vdots \ vec{r}_n} cdot vec{c}_n ,$$
But by assumption, $A^2=0$, meaning all columns of $A^2=0$ and all the above dot products are zero. Hence, $vec{c}_i$ is orthogonal to all vectors in {$vec{r}_1, vec{r}_2 ,...,vec{r}_n $}, and thus all linear combination of those vectors. Since this applies to all $vec{c}_i$, then $W subseteq$ $U^{perp} $.
As for Part B, I'm having trouble and not sure exactly where to begin. Presumably, I would use the results from part A, meaning that if $A^2 ne 0$, then $W notsubseteq$ $U^{perp} $, that is, there is some vector in column space that is not orthogonal to every vector in row space.
Some properties I'm aware of that may be useful but not sure if or how:
$dim(U)=dim(W)$- The orthogonal complement of the row space is null space
- the dimension of the row space and nullspace equals $n$
- dimension of column space of A equals dimension of row space of $A^T$
Anyway, any tips would be greatly appreciated!
linear-algebra proof-writing orthogonality matrix-rank
asked Jan 24 at 7:59
Adam GAdam G
415
$endgroup$
add a comment |
$begingroup$
I'm trying to prepare for an exam and came across the following question:
Given $n times n$ matrix A, let U represent row space and W
represent column space.
A) Prove: $W subseteq$ $U^{perp} $ if and only if $A^2=0$
B) Prove that if $dim(U)>n/2$, then $A^2 ne 0$.
I was able to do part A pretty easily as follows:
$Rightarrow$: Let rows of A be {$ vec{r}_1, vec{r}_2 ,...,vec{r}_n $} and columns be {$ vec{c}_1, vec{c}_2 ,...,vec{c}_n $}. Assume $W subseteq$ $U^{perp} $. Entry $a_{ij}$ (in row i, column j of matrix $A^2=AA$) equals the dot product of $vec{r}_i cdot vec{c}_j $, and by assumption $vec{c}_j$ is orthogonal to $ vec{r}_1, vec{r}_2 ,...,vec{r}_n $, so it follows that $a_{ij}=0$ for all i,j. Thus $A^2=0$
$Leftarrow$: Assume $A^2=0$. We can represent the columns of $A^2$ as
$$ pmatrix{vec{r}_1 \ vec{r}_2 \ vdots \ vec{r}_n} cdot vec{c}_1 , pmatrix{vec{r}_1 \ vec{r}_2 \ vdots \ vec{r}_n} cdot vec{c}_2 , ... , pmatrix{vec{r}_1 \ vec{r}_2 \ vdots \ vec{r}_n} cdot vec{c}_n ,$$
But by assumption, $A^2=0$, meaning all columns of $A^2=0$ and all the above dot products are zero. Hence, $vec{c}_i$ is orthogonal to all vectors in {$vec{r}_1, vec{r}_2 ,...,vec{r}_n $}, and thus all linear combination of those vectors. Since this applies to all $vec{c}_i$, then $W subseteq$ $U^{perp} $.
As for Part B, I'm having trouble and not sure exactly where to begin. Presumably, I would use the results from part A, meaning that if $A^2 ne 0$, then $W notsubseteq$ $U^{perp} $, that is, there is some vector in column space that is not orthogonal to every vector in row space.
Some properties I'm aware of that may be useful but not sure if or how:
$dim(U)=dim(W)$- The orthogonal complement of the row space is null space
- the dimension of the row space and nullspace equals $n$
- dimension of column space of A equals dimension of row space of $A^T$
Anyway, any tips would be greatly appreciated!
linear-algebra proof-writing orthogonality matrix-rank
asked Jan 24 at 7:59
Adam GAdam G
415
$endgroup$
$begingroup$
The assumption in B) is "dim U > n/2". To use the contrapositive, you should assume that $A^2 = 0$, then to prove that $dim U leqslant n/2$.
$endgroup$
– xbh
Jan 24 at 8:04
add a comment |
$begingroup$
I'm trying to prepare for an exam and came across the following question:
Given $n times n$ matrix A, let U represent row space and W
represent column space.
A) Prove: $W subseteq$ $U^{perp} $ if and only if $A^2=0$
B) Prove that if $dim(U)>n/2$, then $A^2 ne 0$.
I was able to do part A pretty easily as follows:
$Rightarrow$: Let rows of A be {$ vec{r}_1, vec{r}_2 ,...,vec{r}_n $} and columns be {$ vec{c}_1, vec{c}_2 ,...,vec{c}_n $}. Assume $W subseteq$ $U^{perp} $. Entry $a_{ij}$ (in row i, column j of matrix $A^2=AA$) equals the dot product of $vec{r}_i cdot vec{c}_j $, and by assumption $vec{c}_j$ is orthogonal to $ vec{r}_1, vec{r}_2 ,...,vec{r}_n $, so it follows that $a_{ij}=0$ for all i,j. Thus $A^2=0$
$Leftarrow$: Assume $A^2=0$. We can represent the columns of $A^2$ as
$$ pmatrix{vec{r}_1 \ vec{r}_2 \ vdots \ vec{r}_n} cdot vec{c}_1 , pmatrix{vec{r}_1 \ vec{r}_2 \ vdots \ vec{r}_n} cdot vec{c}_2 , ... , pmatrix{vec{r}_1 \ vec{r}_2 \ vdots \ vec{r}_n} cdot vec{c}_n ,$$
But by assumption, $A^2=0$, meaning all columns of $A^2=0$ and all the above dot products are zero. Hence, $vec{c}_i$ is orthogonal to all vectors in {$vec{r}_1, vec{r}_2 ,...,vec{r}_n $}, and thus all linear combination of those vectors. Since this applies to all $vec{c}_i$, then $W subseteq$ $U^{perp} $.
As for Part B, I'm having trouble and not sure exactly where to begin. Presumably, I would use the results from part A, meaning that if $A^2 ne 0$, then $W notsubseteq$ $U^{perp} $, that is, there is some vector in column space that is not orthogonal to every vector in row space.
Some properties I'm aware of that may be useful but not sure if or how:
$dim(U)=dim(W)$- The orthogonal complement of the row space is null space
- the dimension of the row space and nullspace equals $n$
- dimension of column space of A equals dimension of row space of $A^T$
Anyway, any tips would be greatly appreciated!
linear-algebra proof-writing orthogonality matrix-rank
asked Jan 24 at 7:59
Adam GAdam G
415
$endgroup$
I'm trying to prepare for an exam and came across the following question:
Given $n times n$ matrix A, let U represent row space and W
represent column space.
A) Prove: $W subseteq$ $U^{perp} $ if and only if $A^2=0$
B) Prove that if $dim(U)>n/2$, then $A^2 ne 0$.
I was able to do part A pretty easily as follows:
$Rightarrow$: Let rows of A be {$ vec{r}_1, vec{r}_2 ,...,vec{r}_n $} and columns be {$ vec{c}_1, vec{c}_2 ,...,vec{c}_n $}. Assume $W subseteq$ $U^{perp} $. Entry $a_{ij}$ (in row i, column j of matrix $A^2=AA$) equals the dot product of $vec{r}_i cdot vec{c}_j $, and by assumption $vec{c}_j$ is orthogonal to $ vec{r}_1, vec{r}_2 ,...,vec{r}_n $, so it follows that $a_{ij}=0$ for all i,j. Thus $A^2=0$
$Leftarrow$: Assume $A^2=0$. We can represent the columns of $A^2$ as
$$ pmatrix{vec{r}_1 \ vec{r}_2 \ vdots \ vec{r}_n} cdot vec{c}_1 , pmatrix{vec{r}_1 \ vec{r}_2 \ vdots \ vec{r}_n} cdot vec{c}_2 , ... , pmatrix{vec{r}_1 \ vec{r}_2 \ vdots \ vec{r}_n} cdot vec{c}_n ,$$
But by assumption, $A^2=0$, meaning all columns of $A^2=0$ and all the above dot products are zero. Hence, $vec{c}_i$ is orthogonal to all vectors in {$vec{r}_1, vec{r}_2 ,...,vec{r}_n $}, and thus all linear combination of those vectors. Since this applies to all $vec{c}_i$, then $W subseteq$ $U^{perp} $.
As for Part B, I'm having trouble and not sure exactly where to begin. Presumably, I would use the results from part A, meaning that if $A^2 ne 0$, then $W notsubseteq$ $U^{perp} $, that is, there is some vector in column space that is not orthogonal to every vector in row space.
Some properties I'm aware of that may be useful but not sure if or how:
$dim(U)=dim(W)$- The orthogonal complement of the row space is null space
- the dimension of the row space and nullspace equals $n$
- dimension of column space of A equals dimension of row space of $A^T$
Anyway, any tips would be greatly appreciated!
linear-algebra proof-writing orthogonality matrix-rank
linear-algebra proof-writing orthogonality matrix-rank
asked Jan 24 at 7:59
Adam GAdam G
415
asked Jan 24 at 7:59
Adam GAdam G
415
asked Jan 24 at 7:59
Adam GAdam G
415
asked Jan 24 at 7:59
Adam GAdam G
415
asked Jan 24 at 7:59
Adam GAdam G
415
415
$begingroup$
The assumption in B) is "dim U > n/2". To use the contrapositive, you should assume that $A^2 = 0$, then to prove that $dim U leqslant n/2$.
$endgroup$
– xbh
Jan 24 at 8:04
add a comment |
$begingroup$
The assumption in B) is "dim U > n/2". To use the contrapositive, you should assume that $A^2 = 0$, then to prove that $dim U leqslant n/2$.
$endgroup$
– xbh
Jan 24 at 8:04
$begingroup$
The assumption in B) is "dim U > n/2". To use the contrapositive, you should assume that $A^2 = 0$, then to prove that $dim U leqslant n/2$.
$endgroup$
– xbh
Jan 24 at 8:04
$begingroup$
The assumption in B) is "dim U > n/2". To use the contrapositive, you should assume that $A^2 = 0$, then to prove that $dim U leqslant n/2$.
$endgroup$
– xbh
Jan 24 at 8:04
add a comment |
2 Answers
2
active
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$begingroup$
Let us regard a matrix as a linear transformation.
$A^2=0$ means Range of $A$ is contained in kernel of $A$.
($A^2v=0$ means $A(Av)=0$ ie $Avin ker A$.)
This is not possible as the dimension of range of $A$ is the dimension of column (row) space of $A$, which is given to be $>n/2$, and kernel dimension, by rank-nullity theorem, is $<n/2$.
answered Jan 24 at 8:24
P VanchinathanP Vanchinathan
15.5k12136
$endgroup$
add a comment |
$begingroup$
Hint: compare the dimensions of $W$ and $U^{perp}$, given that $dim(U) >frac n2$. We can't include a vector space inside one of smaller dimension...
answered Jan 24 at 8:06
jmerryjmerry
14.3k1629
$endgroup$
$begingroup$
Is it as simple as this?: assume $A^2=0$, so from part A, $W subseteq U^{perp}$. Given $dim(U)>n/2$, the dimension of $U^{perp} le n/2$, and therefore $dim(W) le n/2$. But since $dim(U)=dim(W)$, we have $dim(U)>n/2$ and $dim(U) le n/2$, a contradiction. Thus our initial assumption is false.
$endgroup$
– Adam G
Jan 24 at 8:25
$begingroup$
Yes, it's that simple.
$endgroup$
– jmerry
Jan 24 at 8:38
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Let us regard a matrix as a linear transformation.
$A^2=0$ means Range of $A$ is contained in kernel of $A$.
($A^2v=0$ means $A(Av)=0$ ie $Avin ker A$.)
This is not possible as the dimension of range of $A$ is the dimension of column (row) space of $A$, which is given to be $>n/2$, and kernel dimension, by rank-nullity theorem, is $<n/2$.
answered Jan 24 at 8:24
P VanchinathanP Vanchinathan
15.5k12136
$endgroup$
add a comment |
$begingroup$
Let us regard a matrix as a linear transformation.
$A^2=0$ means Range of $A$ is contained in kernel of $A$.
($A^2v=0$ means $A(Av)=0$ ie $Avin ker A$.)
This is not possible as the dimension of range of $A$ is the dimension of column (row) space of $A$, which is given to be $>n/2$, and kernel dimension, by rank-nullity theorem, is $<n/2$.
answered Jan 24 at 8:24
P VanchinathanP Vanchinathan
15.5k12136
$endgroup$
add a comment |
$begingroup$
Let us regard a matrix as a linear transformation.
$A^2=0$ means Range of $A$ is contained in kernel of $A$.
($A^2v=0$ means $A(Av)=0$ ie $Avin ker A$.)
This is not possible as the dimension of range of $A$ is the dimension of column (row) space of $A$, which is given to be $>n/2$, and kernel dimension, by rank-nullity theorem, is $<n/2$.
answered Jan 24 at 8:24
P VanchinathanP Vanchinathan
15.5k12136
$endgroup$
Let us regard a matrix as a linear transformation.
$A^2=0$ means Range of $A$ is contained in kernel of $A$.
($A^2v=0$ means $A(Av)=0$ ie $Avin ker A$.)
This is not possible as the dimension of range of $A$ is the dimension of column (row) space of $A$, which is given to be $>n/2$, and kernel dimension, by rank-nullity theorem, is $<n/2$.
answered Jan 24 at 8:24
P VanchinathanP Vanchinathan
15.5k12136
answered Jan 24 at 8:24
P VanchinathanP Vanchinathan
15.5k12136
answered Jan 24 at 8:24
P VanchinathanP Vanchinathan
15.5k12136
answered Jan 24 at 8:24
P VanchinathanP Vanchinathan
15.5k12136
15.5k12136
add a comment |
add a comment |
$begingroup$
Hint: compare the dimensions of $W$ and $U^{perp}$, given that $dim(U) >frac n2$. We can't include a vector space inside one of smaller dimension...
answered Jan 24 at 8:06
jmerryjmerry
14.3k1629
$endgroup$
$begingroup$
Is it as simple as this?: assume $A^2=0$, so from part A, $W subseteq U^{perp}$. Given $dim(U)>n/2$, the dimension of $U^{perp} le n/2$, and therefore $dim(W) le n/2$. But since $dim(U)=dim(W)$, we have $dim(U)>n/2$ and $dim(U) le n/2$, a contradiction. Thus our initial assumption is false.
$endgroup$
– Adam G
Jan 24 at 8:25
$begingroup$
Yes, it's that simple.
$endgroup$
– jmerry
Jan 24 at 8:38
add a comment |
$begingroup$
Hint: compare the dimensions of $W$ and $U^{perp}$, given that $dim(U) >frac n2$. We can't include a vector space inside one of smaller dimension...
answered Jan 24 at 8:06
jmerryjmerry
14.3k1629
$endgroup$
$begingroup$
Is it as simple as this?: assume $A^2=0$, so from part A, $W subseteq U^{perp}$. Given $dim(U)>n/2$, the dimension of $U^{perp} le n/2$, and therefore $dim(W) le n/2$. But since $dim(U)=dim(W)$, we have $dim(U)>n/2$ and $dim(U) le n/2$, a contradiction. Thus our initial assumption is false.
$endgroup$
– Adam G
Jan 24 at 8:25
$begingroup$
Yes, it's that simple.
$endgroup$
– jmerry
Jan 24 at 8:38
add a comment |
$begingroup$
Hint: compare the dimensions of $W$ and $U^{perp}$, given that $dim(U) >frac n2$. We can't include a vector space inside one of smaller dimension...
answered Jan 24 at 8:06
jmerryjmerry
14.3k1629
$endgroup$
Hint: compare the dimensions of $W$ and $U^{perp}$, given that $dim(U) >frac n2$. We can't include a vector space inside one of smaller dimension...
answered Jan 24 at 8:06
jmerryjmerry
14.3k1629
answered Jan 24 at 8:06
jmerryjmerry
14.3k1629
answered Jan 24 at 8:06
jmerryjmerry
14.3k1629
answered Jan 24 at 8:06
jmerryjmerry
14.3k1629
14.3k1629
$begingroup$
Is it as simple as this?: assume $A^2=0$, so from part A, $W subseteq U^{perp}$. Given $dim(U)>n/2$, the dimension of $U^{perp} le n/2$, and therefore $dim(W) le n/2$. But since $dim(U)=dim(W)$, we have $dim(U)>n/2$ and $dim(U) le n/2$, a contradiction. Thus our initial assumption is false.
$endgroup$
– Adam G
Jan 24 at 8:25
$begingroup$
Yes, it's that simple.
$endgroup$
– jmerry
Jan 24 at 8:38
add a comment |
$begingroup$
Is it as simple as this?: assume $A^2=0$, so from part A, $W subseteq U^{perp}$. Given $dim(U)>n/2$, the dimension of $U^{perp} le n/2$, and therefore $dim(W) le n/2$. But since $dim(U)=dim(W)$, we have $dim(U)>n/2$ and $dim(U) le n/2$, a contradiction. Thus our initial assumption is false.
$endgroup$
– Adam G
Jan 24 at 8:25
$begingroup$
Yes, it's that simple.
$endgroup$
– jmerry
Jan 24 at 8:38
$begingroup$
Is it as simple as this?: assume $A^2=0$, so from part A, $W subseteq U^{perp}$. Given $dim(U)>n/2$, the dimension of $U^{perp} le n/2$, and therefore $dim(W) le n/2$. But since $dim(U)=dim(W)$, we have $dim(U)>n/2$ and $dim(U) le n/2$, a contradiction. Thus our initial assumption is false.
$endgroup$
– Adam G
Jan 24 at 8:25
$begingroup$
Is it as simple as this?: assume $A^2=0$, so from part A, $W subseteq U^{perp}$. Given $dim(U)>n/2$, the dimension of $U^{perp} le n/2$, and therefore $dim(W) le n/2$. But since $dim(U)=dim(W)$, we have $dim(U)>n/2$ and $dim(U) le n/2$, a contradiction. Thus our initial assumption is false.
$endgroup$
– Adam G
Jan 24 at 8:25
$begingroup$
Yes, it's that simple.
$endgroup$
– jmerry
Jan 24 at 8:38
$begingroup$
Yes, it's that simple.
$endgroup$
– jmerry
Jan 24 at 8:38
add a comment |
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$begingroup$
The assumption in B) is "dim U > n/2". To use the contrapositive, you should assume that $A^2 = 0$, then to prove that $dim U leqslant n/2$.
$endgroup$
– xbh
Jan 24 at 8:04