Show that if $B^2 x = 0_n$ for some vector $x neq 0_n$, then $B$ is not invertible
$begingroup$
If $B in M_{n times n}(mathbb{R})$ and $B^2 x = 0_n$ for some vector $x neq 0_n$, then $B$ is not invertible.
I get that $$mbox{rank} ( B^2 ) < n$$ but I can't seem to be able to link it to $B$ . Perhaps I need to use diagonalization to deal with the power, but that only works if $B$ is diagonalizable. Any hints would be appreciated.
linear-algebra matrices
edited Aug 1 '17 at 11:32
Rodrigo de Azevedo
13.1k41960
asked Aug 1 '17 at 1:18
MigosMigos
1014
$endgroup$
add a comment |
$begingroup$
If $B in M_{n times n}(mathbb{R})$ and $B^2 x = 0_n$ for some vector $x neq 0_n$, then $B$ is not invertible.
I get that $$mbox{rank} ( B^2 ) < n$$ but I can't seem to be able to link it to $B$ . Perhaps I need to use diagonalization to deal with the power, but that only works if $B$ is diagonalizable. Any hints would be appreciated.
linear-algebra matrices
edited Aug 1 '17 at 11:32
Rodrigo de Azevedo
13.1k41960
asked Aug 1 '17 at 1:18
MigosMigos
1014
$endgroup$
add a comment |
$begingroup$
If $B in M_{n times n}(mathbb{R})$ and $B^2 x = 0_n$ for some vector $x neq 0_n$, then $B$ is not invertible.
I get that $$mbox{rank} ( B^2 ) < n$$ but I can't seem to be able to link it to $B$ . Perhaps I need to use diagonalization to deal with the power, but that only works if $B$ is diagonalizable. Any hints would be appreciated.
linear-algebra matrices
edited Aug 1 '17 at 11:32
Rodrigo de Azevedo
13.1k41960
asked Aug 1 '17 at 1:18
MigosMigos
1014
$endgroup$
If $B in M_{n times n}(mathbb{R})$ and $B^2 x = 0_n$ for some vector $x neq 0_n$, then $B$ is not invertible.
I get that $$mbox{rank} ( B^2 ) < n$$ but I can't seem to be able to link it to $B$ . Perhaps I need to use diagonalization to deal with the power, but that only works if $B$ is diagonalizable. Any hints would be appreciated.
linear-algebra matrices
linear-algebra matrices
edited Aug 1 '17 at 11:32
Rodrigo de Azevedo
13.1k41960
asked Aug 1 '17 at 1:18
MigosMigos
1014
edited Aug 1 '17 at 11:32
Rodrigo de Azevedo
13.1k41960
asked Aug 1 '17 at 1:18
MigosMigos
1014
edited Aug 1 '17 at 11:32
Rodrigo de Azevedo
13.1k41960
edited Aug 1 '17 at 11:32
Rodrigo de Azevedo
13.1k41960
edited Aug 1 '17 at 11:32
Rodrigo de Azevedo
13.1k41960
13.1k41960
asked Aug 1 '17 at 1:18
MigosMigos
1014
asked Aug 1 '17 at 1:18
MigosMigos
1014
asked Aug 1 '17 at 1:18
MigosMigos
1014
1014
add a comment |
add a comment |
3 Answers
3
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oldest
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$begingroup$
A linear transformation in finite dimensional space is injective iff its kernel is trivial. So we have to find a non-trivial element in the kernel.
If $B^2 x = 0$, then $B(Bx) = 0$, so $Bx in ker B$.
Suppose $Bx = 0$, then this shows that $B$ has a non-trivial kernel, hence is not injective.
Suppose $Bx neq 0$ above, then $Bx$ is a non-trivial element of $ker B$, so it is not injective for $n geq 3$.
Hence, either way it follows that $B$ is not injective.
VIA CONTRAPOSITIVE : If $B$ is injective, then so is $B^2$, but then $B^20 = 0$, so $B^2x$ cannot be zero for any other $x$ by injectivity. Note that we can extend this to $B^nx = 0$ implies $B$ is not injective.
answered Aug 1 '17 at 1:26
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.6k33678
$endgroup$
add a comment |
$begingroup$
If $B^2x=0$ for some non-zero vector $x$, then $B^2$ is not invertible, so $det(B^2)=0$.
But $det(B^2)=det(B)^2$, so $det(B)=0$ and $B$ is not invertible.
answered Aug 1 '17 at 1:26
carmichael561carmichael561
47.2k54583
$endgroup$
add a comment |
$begingroup$
Another way to look at this is by contrapositive: suppose $B$ is invertible. Then $B^2$ is invertible, so $B^2 x = 0$ only for $x=0$.
answered Aug 1 '17 at 4:09
Bob KruegerBob Krueger
4,2002822
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3 Answers
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3 Answers
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$begingroup$
A linear transformation in finite dimensional space is injective iff its kernel is trivial. So we have to find a non-trivial element in the kernel.
If $B^2 x = 0$, then $B(Bx) = 0$, so $Bx in ker B$.
Suppose $Bx = 0$, then this shows that $B$ has a non-trivial kernel, hence is not injective.
Suppose $Bx neq 0$ above, then $Bx$ is a non-trivial element of $ker B$, so it is not injective for $n geq 3$.
Hence, either way it follows that $B$ is not injective.
VIA CONTRAPOSITIVE : If $B$ is injective, then so is $B^2$, but then $B^20 = 0$, so $B^2x$ cannot be zero for any other $x$ by injectivity. Note that we can extend this to $B^nx = 0$ implies $B$ is not injective.
answered Aug 1 '17 at 1:26
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.6k33678
$endgroup$
add a comment |
$begingroup$
A linear transformation in finite dimensional space is injective iff its kernel is trivial. So we have to find a non-trivial element in the kernel.
If $B^2 x = 0$, then $B(Bx) = 0$, so $Bx in ker B$.
Suppose $Bx = 0$, then this shows that $B$ has a non-trivial kernel, hence is not injective.
Suppose $Bx neq 0$ above, then $Bx$ is a non-trivial element of $ker B$, so it is not injective for $n geq 3$.
Hence, either way it follows that $B$ is not injective.
VIA CONTRAPOSITIVE : If $B$ is injective, then so is $B^2$, but then $B^20 = 0$, so $B^2x$ cannot be zero for any other $x$ by injectivity. Note that we can extend this to $B^nx = 0$ implies $B$ is not injective.
answered Aug 1 '17 at 1:26
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.6k33678
$endgroup$
add a comment |
$begingroup$
A linear transformation in finite dimensional space is injective iff its kernel is trivial. So we have to find a non-trivial element in the kernel.
If $B^2 x = 0$, then $B(Bx) = 0$, so $Bx in ker B$.
Suppose $Bx = 0$, then this shows that $B$ has a non-trivial kernel, hence is not injective.
Suppose $Bx neq 0$ above, then $Bx$ is a non-trivial element of $ker B$, so it is not injective for $n geq 3$.
Hence, either way it follows that $B$ is not injective.
VIA CONTRAPOSITIVE : If $B$ is injective, then so is $B^2$, but then $B^20 = 0$, so $B^2x$ cannot be zero for any other $x$ by injectivity. Note that we can extend this to $B^nx = 0$ implies $B$ is not injective.
answered Aug 1 '17 at 1:26
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.6k33678
$endgroup$
A linear transformation in finite dimensional space is injective iff its kernel is trivial. So we have to find a non-trivial element in the kernel.
If $B^2 x = 0$, then $B(Bx) = 0$, so $Bx in ker B$.
Suppose $Bx = 0$, then this shows that $B$ has a non-trivial kernel, hence is not injective.
Suppose $Bx neq 0$ above, then $Bx$ is a non-trivial element of $ker B$, so it is not injective for $n geq 3$.
Hence, either way it follows that $B$ is not injective.
VIA CONTRAPOSITIVE : If $B$ is injective, then so is $B^2$, but then $B^20 = 0$, so $B^2x$ cannot be zero for any other $x$ by injectivity. Note that we can extend this to $B^nx = 0$ implies $B$ is not injective.
answered Aug 1 '17 at 1:26
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.6k33678
edited Jan 8 at 16:50
answered Aug 1 '17 at 1:26
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.6k33678
answered Aug 1 '17 at 1:26
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.6k33678
answered Aug 1 '17 at 1:26
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.6k33678
40.6k33678
add a comment |
add a comment |
$begingroup$
If $B^2x=0$ for some non-zero vector $x$, then $B^2$ is not invertible, so $det(B^2)=0$.
But $det(B^2)=det(B)^2$, so $det(B)=0$ and $B$ is not invertible.
answered Aug 1 '17 at 1:26
carmichael561carmichael561
47.2k54583
$endgroup$
add a comment |
$begingroup$
If $B^2x=0$ for some non-zero vector $x$, then $B^2$ is not invertible, so $det(B^2)=0$.
But $det(B^2)=det(B)^2$, so $det(B)=0$ and $B$ is not invertible.
answered Aug 1 '17 at 1:26
carmichael561carmichael561
47.2k54583
$endgroup$
add a comment |
$begingroup$
If $B^2x=0$ for some non-zero vector $x$, then $B^2$ is not invertible, so $det(B^2)=0$.
But $det(B^2)=det(B)^2$, so $det(B)=0$ and $B$ is not invertible.
answered Aug 1 '17 at 1:26
carmichael561carmichael561
47.2k54583
$endgroup$
If $B^2x=0$ for some non-zero vector $x$, then $B^2$ is not invertible, so $det(B^2)=0$.
But $det(B^2)=det(B)^2$, so $det(B)=0$ and $B$ is not invertible.
answered Aug 1 '17 at 1:26
carmichael561carmichael561
47.2k54583
answered Aug 1 '17 at 1:26
carmichael561carmichael561
47.2k54583
answered Aug 1 '17 at 1:26
carmichael561carmichael561
47.2k54583
answered Aug 1 '17 at 1:26
carmichael561carmichael561
47.2k54583
47.2k54583
add a comment |
add a comment |
$begingroup$
Another way to look at this is by contrapositive: suppose $B$ is invertible. Then $B^2$ is invertible, so $B^2 x = 0$ only for $x=0$.
answered Aug 1 '17 at 4:09
Bob KruegerBob Krueger
4,2002822
$endgroup$
add a comment |
$begingroup$
Another way to look at this is by contrapositive: suppose $B$ is invertible. Then $B^2$ is invertible, so $B^2 x = 0$ only for $x=0$.
answered Aug 1 '17 at 4:09
Bob KruegerBob Krueger
4,2002822
$endgroup$
add a comment |
$begingroup$
Another way to look at this is by contrapositive: suppose $B$ is invertible. Then $B^2$ is invertible, so $B^2 x = 0$ only for $x=0$.
answered Aug 1 '17 at 4:09
Bob KruegerBob Krueger
4,2002822
$endgroup$
Another way to look at this is by contrapositive: suppose $B$ is invertible. Then $B^2$ is invertible, so $B^2 x = 0$ only for $x=0$.
answered Aug 1 '17 at 4:09
Bob KruegerBob Krueger
4,2002822
answered Aug 1 '17 at 4:09
Bob KruegerBob Krueger
4,2002822
answered Aug 1 '17 at 4:09
Bob KruegerBob Krueger
4,2002822
answered Aug 1 '17 at 4:09
Bob KruegerBob Krueger
4,2002822
4,2002822
add a comment |
add a comment |
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