Regarding a real $n times n$ matrix $A$ satisfying $A^6 = -A^2$
$begingroup$
The Problem:
Suppose $A$ represents a real $n times n$ matrix satisfying $A^6 = -A^2$.
(a) Prove that if $A$ is symmetric, then $A = 0$.
(b) Prove that if $n$ is odd, then $A$ is not invertible.
(c) Give an example of an invertible $A$ with $n$ even (e.g., $n = 2$).
My Approach:
For (a): I'm just going through all of the properties of real symmetric matrices and seeing if anything pops out. Let $lambda_1, ..., lambda_n$ be the eigenvalues of $A$. Then we have that $A = QDQ^{-1}$, where $D$ is diagonal (with $lambda_1, ..., lambda_n$ as its diagonal entries) and $Q$ is a real orthogonal matrix (with the eigenvectors of $A$ as its columns). So, since $A^6 = -A^2$, it follows that $D^6 = -D^2$; and so $lambda_i^6 = -(lambda_i^2)$, for each $i in {1,2,...,n}$. But since $lambda_1, ..., lambda_n in mathbb{R}$, it must then be that $lambda_1 = cdots =lambda_n = 0$. Well, that means that $D = 0$, but that doesn't necessarily mean that $A = 0$... or does it?
For (b): (I think I've got this one.) Since $A^6 = -A^2$, it follows that $text{det}(A)^6 = (-1)^ntext{det}(A)^2$. But the LHS of this equation is positive and the RHS is negative (since $n$ is odd). Thus $text{det}(A)$ must be $0$; and so $A$ is not invertible.
For (c): I can easily come up with examples satisfying $A^6 = A^2$; e.g.,
begin{pmatrix}
0 & 1 \
-1 & 0 \
end{pmatrix}
But that minus sign is killing me. Of course, I'm just using trial-and-error here; I'd certainly prefer a more "informed" approach to finding such a matrix...
linear-algebra matrices matrix-equations
edited Jan 13 at 17:02
José Carlos Santos
168k23132236
asked Jan 13 at 16:50
thisisourconcerndudethisisourconcerndude
1,1271123
$endgroup$
add a comment |
$begingroup$
The Problem:
Suppose $A$ represents a real $n times n$ matrix satisfying $A^6 = -A^2$.
(a) Prove that if $A$ is symmetric, then $A = 0$.
(b) Prove that if $n$ is odd, then $A$ is not invertible.
(c) Give an example of an invertible $A$ with $n$ even (e.g., $n = 2$).
My Approach:
For (a): I'm just going through all of the properties of real symmetric matrices and seeing if anything pops out. Let $lambda_1, ..., lambda_n$ be the eigenvalues of $A$. Then we have that $A = QDQ^{-1}$, where $D$ is diagonal (with $lambda_1, ..., lambda_n$ as its diagonal entries) and $Q$ is a real orthogonal matrix (with the eigenvectors of $A$ as its columns). So, since $A^6 = -A^2$, it follows that $D^6 = -D^2$; and so $lambda_i^6 = -(lambda_i^2)$, for each $i in {1,2,...,n}$. But since $lambda_1, ..., lambda_n in mathbb{R}$, it must then be that $lambda_1 = cdots =lambda_n = 0$. Well, that means that $D = 0$, but that doesn't necessarily mean that $A = 0$... or does it?
For (b): (I think I've got this one.) Since $A^6 = -A^2$, it follows that $text{det}(A)^6 = (-1)^ntext{det}(A)^2$. But the LHS of this equation is positive and the RHS is negative (since $n$ is odd). Thus $text{det}(A)$ must be $0$; and so $A$ is not invertible.
For (c): I can easily come up with examples satisfying $A^6 = A^2$; e.g.,
begin{pmatrix}
0 & 1 \
-1 & 0 \
end{pmatrix}
But that minus sign is killing me. Of course, I'm just using trial-and-error here; I'd certainly prefer a more "informed" approach to finding such a matrix...
linear-algebra matrices matrix-equations
edited Jan 13 at 17:02
José Carlos Santos
168k23132236
asked Jan 13 at 16:50
thisisourconcerndudethisisourconcerndude
1,1271123
$endgroup$
add a comment |
$begingroup$
The Problem:
Suppose $A$ represents a real $n times n$ matrix satisfying $A^6 = -A^2$.
(a) Prove that if $A$ is symmetric, then $A = 0$.
(b) Prove that if $n$ is odd, then $A$ is not invertible.
(c) Give an example of an invertible $A$ with $n$ even (e.g., $n = 2$).
My Approach:
For (a): I'm just going through all of the properties of real symmetric matrices and seeing if anything pops out. Let $lambda_1, ..., lambda_n$ be the eigenvalues of $A$. Then we have that $A = QDQ^{-1}$, where $D$ is diagonal (with $lambda_1, ..., lambda_n$ as its diagonal entries) and $Q$ is a real orthogonal matrix (with the eigenvectors of $A$ as its columns). So, since $A^6 = -A^2$, it follows that $D^6 = -D^2$; and so $lambda_i^6 = -(lambda_i^2)$, for each $i in {1,2,...,n}$. But since $lambda_1, ..., lambda_n in mathbb{R}$, it must then be that $lambda_1 = cdots =lambda_n = 0$. Well, that means that $D = 0$, but that doesn't necessarily mean that $A = 0$... or does it?
For (b): (I think I've got this one.) Since $A^6 = -A^2$, it follows that $text{det}(A)^6 = (-1)^ntext{det}(A)^2$. But the LHS of this equation is positive and the RHS is negative (since $n$ is odd). Thus $text{det}(A)$ must be $0$; and so $A$ is not invertible.
For (c): I can easily come up with examples satisfying $A^6 = A^2$; e.g.,
begin{pmatrix}
0 & 1 \
-1 & 0 \
end{pmatrix}
But that minus sign is killing me. Of course, I'm just using trial-and-error here; I'd certainly prefer a more "informed" approach to finding such a matrix...
linear-algebra matrices matrix-equations
edited Jan 13 at 17:02
José Carlos Santos
168k23132236
asked Jan 13 at 16:50
thisisourconcerndudethisisourconcerndude
1,1271123
$endgroup$
The Problem:
Suppose $A$ represents a real $n times n$ matrix satisfying $A^6 = -A^2$.
(a) Prove that if $A$ is symmetric, then $A = 0$.
(b) Prove that if $n$ is odd, then $A$ is not invertible.
(c) Give an example of an invertible $A$ with $n$ even (e.g., $n = 2$).
My Approach:
For (a): I'm just going through all of the properties of real symmetric matrices and seeing if anything pops out. Let $lambda_1, ..., lambda_n$ be the eigenvalues of $A$. Then we have that $A = QDQ^{-1}$, where $D$ is diagonal (with $lambda_1, ..., lambda_n$ as its diagonal entries) and $Q$ is a real orthogonal matrix (with the eigenvectors of $A$ as its columns). So, since $A^6 = -A^2$, it follows that $D^6 = -D^2$; and so $lambda_i^6 = -(lambda_i^2)$, for each $i in {1,2,...,n}$. But since $lambda_1, ..., lambda_n in mathbb{R}$, it must then be that $lambda_1 = cdots =lambda_n = 0$. Well, that means that $D = 0$, but that doesn't necessarily mean that $A = 0$... or does it?
For (b): (I think I've got this one.) Since $A^6 = -A^2$, it follows that $text{det}(A)^6 = (-1)^ntext{det}(A)^2$. But the LHS of this equation is positive and the RHS is negative (since $n$ is odd). Thus $text{det}(A)$ must be $0$; and so $A$ is not invertible.
For (c): I can easily come up with examples satisfying $A^6 = A^2$; e.g.,
begin{pmatrix}
0 & 1 \
-1 & 0 \
end{pmatrix}
But that minus sign is killing me. Of course, I'm just using trial-and-error here; I'd certainly prefer a more "informed" approach to finding such a matrix...
linear-algebra matrices matrix-equations
linear-algebra matrices matrix-equations
edited Jan 13 at 17:02
José Carlos Santos
168k23132236
asked Jan 13 at 16:50
thisisourconcerndudethisisourconcerndude
1,1271123
edited Jan 13 at 17:02
José Carlos Santos
168k23132236
asked Jan 13 at 16:50
thisisourconcerndudethisisourconcerndude
1,1271123
edited Jan 13 at 17:02
José Carlos Santos
168k23132236
edited Jan 13 at 17:02
José Carlos Santos
168k23132236
edited Jan 13 at 17:02
José Carlos Santos
168k23132236
168k23132236
asked Jan 13 at 16:50
thisisourconcerndudethisisourconcerndude
1,1271123
asked Jan 13 at 16:50
thisisourconcerndudethisisourconcerndude
1,1271123
asked Jan 13 at 16:50
thisisourconcerndudethisisourconcerndude
1,1271123
1,1271123
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
(a) Since $A$ and $D$ are similar matrices, $D=0$ if and only if $A=0$.
(b) Right.
(c) Let us consider the following equation in $mathbb C$: $z^6=-z^2$. You have$$z^6=-z^2iff z=0vee z^4=-1.$$So, take a fourth root of $-1$, such as $cosleft(fracpi4right)+isinleft(fracpi4right)$. Now, write $z$ as a $2times2$ matrix:$$z=begin{bmatrix}frac{sqrt2}2&-frac{sqrt2}2\frac{sqrt2}2&frac{sqrt2}2end{bmatrix}.$$This approach works because$$begin{array}{ccc}mathbb C&longrightarrow&mathbb{R}^{2times2}\a+bi&mapsto&begin{bmatrix}a&-b\b&aend{bmatrix}end{array}$$is an injective ring homomorphism.
answered Jan 13 at 16:57
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
(a) Since $A$ and $D$ are similar matrices, $D=0$ if and only if $A=0$.
(b) Right.
(c) Let us consider the following equation in $mathbb C$: $z^6=-z^2$. You have$$z^6=-z^2iff z=0vee z^4=-1.$$So, take a fourth root of $-1$, such as $cosleft(fracpi4right)+isinleft(fracpi4right)$. Now, write $z$ as a $2times2$ matrix:$$z=begin{bmatrix}frac{sqrt2}2&-frac{sqrt2}2\frac{sqrt2}2&frac{sqrt2}2end{bmatrix}.$$This approach works because$$begin{array}{ccc}mathbb C&longrightarrow&mathbb{R}^{2times2}\a+bi&mapsto&begin{bmatrix}a&-b\b&aend{bmatrix}end{array}$$is an injective ring homomorphism.
answered Jan 13 at 16:57
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
add a comment |
$begingroup$
(a) Since $A$ and $D$ are similar matrices, $D=0$ if and only if $A=0$.
(b) Right.
(c) Let us consider the following equation in $mathbb C$: $z^6=-z^2$. You have$$z^6=-z^2iff z=0vee z^4=-1.$$So, take a fourth root of $-1$, such as $cosleft(fracpi4right)+isinleft(fracpi4right)$. Now, write $z$ as a $2times2$ matrix:$$z=begin{bmatrix}frac{sqrt2}2&-frac{sqrt2}2\frac{sqrt2}2&frac{sqrt2}2end{bmatrix}.$$This approach works because$$begin{array}{ccc}mathbb C&longrightarrow&mathbb{R}^{2times2}\a+bi&mapsto&begin{bmatrix}a&-b\b&aend{bmatrix}end{array}$$is an injective ring homomorphism.
answered Jan 13 at 16:57
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
add a comment |
$begingroup$
(a) Since $A$ and $D$ are similar matrices, $D=0$ if and only if $A=0$.
(b) Right.
(c) Let us consider the following equation in $mathbb C$: $z^6=-z^2$. You have$$z^6=-z^2iff z=0vee z^4=-1.$$So, take a fourth root of $-1$, such as $cosleft(fracpi4right)+isinleft(fracpi4right)$. Now, write $z$ as a $2times2$ matrix:$$z=begin{bmatrix}frac{sqrt2}2&-frac{sqrt2}2\frac{sqrt2}2&frac{sqrt2}2end{bmatrix}.$$This approach works because$$begin{array}{ccc}mathbb C&longrightarrow&mathbb{R}^{2times2}\a+bi&mapsto&begin{bmatrix}a&-b\b&aend{bmatrix}end{array}$$is an injective ring homomorphism.
answered Jan 13 at 16:57
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
(a) Since $A$ and $D$ are similar matrices, $D=0$ if and only if $A=0$.
(b) Right.
(c) Let us consider the following equation in $mathbb C$: $z^6=-z^2$. You have$$z^6=-z^2iff z=0vee z^4=-1.$$So, take a fourth root of $-1$, such as $cosleft(fracpi4right)+isinleft(fracpi4right)$. Now, write $z$ as a $2times2$ matrix:$$z=begin{bmatrix}frac{sqrt2}2&-frac{sqrt2}2\frac{sqrt2}2&frac{sqrt2}2end{bmatrix}.$$This approach works because$$begin{array}{ccc}mathbb C&longrightarrow&mathbb{R}^{2times2}\a+bi&mapsto&begin{bmatrix}a&-b\b&aend{bmatrix}end{array}$$is an injective ring homomorphism.
answered Jan 13 at 16:57
José Carlos SantosJosé Carlos Santos
168k23132236
edited Jan 13 at 17:11
answered Jan 13 at 16:57
José Carlos SantosJosé Carlos Santos
168k23132236
answered Jan 13 at 16:57
José Carlos SantosJosé Carlos Santos
168k23132236
answered Jan 13 at 16:57
José Carlos SantosJosé Carlos Santos
168k23132236
168k23132236
add a comment |
add a comment |
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