Mixing
Relation between trace of matrix $A^*A$ and invertibility of the matrix $A$
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Is there any relation between $mathrm{tr}(A^*A)$ and invertibility of the matrix $A$? What information about the matrix $A$ does $A^*A$ gives ?
I was confused about this when I came across the following statements:
Is it true that :
- If $A$ is invertible, then $mathrm{tr}(A^*A)$ is non zero.
- If $|mathrm{tr}(A^*A)|<n^2$, then $|a_{ij}|<1$ for some $i,j$
- If $|mathrm{tr}(A^*A)|=0$, then $A$ is a zero matrix.
Can someone help me with the answer? Thank you in advance.
linear-algebra trace
edited yesterday
Jean-Claude Arbaut
14.3k63361
asked yesterday
math math
84
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math math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
up vote
1
down vote
favorite
Is there any relation between $mathrm{tr}(A^*A)$ and invertibility of the matrix $A$? What information about the matrix $A$ does $A^*A$ gives ?
I was confused about this when I came across the following statements:
Is it true that :
- If $A$ is invertible, then $mathrm{tr}(A^*A)$ is non zero.
- If $|mathrm{tr}(A^*A)|<n^2$, then $|a_{ij}|<1$ for some $i,j$
- If $|mathrm{tr}(A^*A)|=0$, then $A$ is a zero matrix.
Can someone help me with the answer? Thank you in advance.
linear-algebra trace
edited yesterday
Jean-Claude Arbaut
14.3k63361
asked yesterday
math math
84
New contributor
math math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What is $A^*$ in context? Complex conjugate of $A$? Conjugate-transpose of $A$?
– Richard Martin
yesterday
@RichardMartin Conjugate transpose
– Jean-Claude Arbaut
yesterday
3
Hint: $mathrm{tr}(A^*A)=sum_{i,j} |a_{ij}|^2$. @RichardMartin No, it's not the "sum of squares", there is a modulus.
– Jean-Claude Arbaut
yesterday
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Is there any relation between $mathrm{tr}(A^*A)$ and invertibility of the matrix $A$? What information about the matrix $A$ does $A^*A$ gives ?
I was confused about this when I came across the following statements:
Is it true that :
- If $A$ is invertible, then $mathrm{tr}(A^*A)$ is non zero.
- If $|mathrm{tr}(A^*A)|<n^2$, then $|a_{ij}|<1$ for some $i,j$
- If $|mathrm{tr}(A^*A)|=0$, then $A$ is a zero matrix.
Can someone help me with the answer? Thank you in advance.
linear-algebra trace
edited yesterday
Jean-Claude Arbaut
14.3k63361
asked yesterday
math math
84
New contributor
math math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Is there any relation between $mathrm{tr}(A^*A)$ and invertibility of the matrix $A$? What information about the matrix $A$ does $A^*A$ gives ?
I was confused about this when I came across the following statements:
Is it true that :
- If $A$ is invertible, then $mathrm{tr}(A^*A)$ is non zero.
- If $|mathrm{tr}(A^*A)|<n^2$, then $|a_{ij}|<1$ for some $i,j$
- If $|mathrm{tr}(A^*A)|=0$, then $A$ is a zero matrix.
Can someone help me with the answer? Thank you in advance.
linear-algebra trace
linear-algebra trace
edited yesterday
Jean-Claude Arbaut
14.3k63361
asked yesterday
math math
84
New contributor
math math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Jean-Claude Arbaut
14.3k63361
asked yesterday
math math
84
New contributor
math math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Jean-Claude Arbaut
14.3k63361
edited yesterday
Jean-Claude Arbaut
14.3k63361
edited yesterday
Jean-Claude Arbaut
14.3k63361
14.3k63361
asked yesterday
math math
84
New contributor
math math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
math math
84
asked yesterday
math math
84
84
New contributor
math math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
math math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
math math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What is $A^*$ in context? Complex conjugate of $A$? Conjugate-transpose of $A$?
– Richard Martin
yesterday
@RichardMartin Conjugate transpose
– Jean-Claude Arbaut
yesterday
3
Hint: $mathrm{tr}(A^*A)=sum_{i,j} |a_{ij}|^2$. @RichardMartin No, it's not the "sum of squares", there is a modulus.
– Jean-Claude Arbaut
yesterday
add a comment |
What is $A^*$ in context? Complex conjugate of $A$? Conjugate-transpose of $A$?
– Richard Martin
yesterday
@RichardMartin Conjugate transpose
– Jean-Claude Arbaut
yesterday
3
Hint: $mathrm{tr}(A^*A)=sum_{i,j} |a_{ij}|^2$. @RichardMartin No, it's not the "sum of squares", there is a modulus.
– Jean-Claude Arbaut
yesterday
What is $A^*$ in context? Complex conjugate of $A$? Conjugate-transpose of $A$?
– Richard Martin
yesterday
What is $A^*$ in context? Complex conjugate of $A$? Conjugate-transpose of $A$?
– Richard Martin
yesterday
@RichardMartin Conjugate transpose
– Jean-Claude Arbaut
yesterday
@RichardMartin Conjugate transpose
– Jean-Claude Arbaut
yesterday
3
3
Hint: $mathrm{tr}(A^*A)=sum_{i,j} |a_{ij}|^2$. @RichardMartin No, it's not the "sum of squares", there is a modulus.
– Jean-Claude Arbaut
yesterday
Hint: $mathrm{tr}(A^*A)=sum_{i,j} |a_{ij}|^2$. @RichardMartin No, it's not the "sum of squares", there is a modulus.
– Jean-Claude Arbaut
yesterday
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Let $A in mathbb{C}^{n times n}$ be a complex square matrix.
Since for a complex number $z = x + iy$ we have
$$
z^* z = (x - iy)(x + iy) = x^2 + y^2 = |z|^2,
quad text{where } |z| := sqrt{x^2 + y^2}
$$
Now we have
$$
A^* A =
begin{pmatrix}
a_{1,1}^* & ldots & a_{n,1}^* \
vdots & ddots & vdots \
a_{1,n}^* & ldots & a_{n,n}^*
end{pmatrix}
begin{pmatrix}
a_{1,1} & ldots & a_{1,n} \
vdots & ddots & vdots \
a_{n,1} & ldots & a_{n,n}
end{pmatrix}
= begin{pmatrix}
sum_{i = 1}^{n} a_{i,1} a_{i,1}^* & & ast \
& ddots & \
ast & & sum_{i = 1}^{n} a_{i,n} a_{i,n}^*
end{pmatrix}
$$
therefore we have $operatorname{tr}(A^* A) = sum_{i,j=1}^{n} |a_{i,j}|^2$
- Let let $operatorname{tr}(A^*A) = 0$ then $|a_{i,j}|=0$ for all so the $A$ is the zero matrix, so $det(A)=0$, so it's not invertible, proving (1) by contraposition.
- From the above sum formula for the trace, since all summands are positive, when the sum is smaller than $n^2$, which is the number of summands, one of the summands has to be smaller than 1.
- Proven in (1).
answered yesterday
Viktor Glombik
474220
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $A in mathbb{C}^{n times n}$ be a complex square matrix.
Since for a complex number $z = x + iy$ we have
$$
z^* z = (x - iy)(x + iy) = x^2 + y^2 = |z|^2,
quad text{where } |z| := sqrt{x^2 + y^2}
$$
Now we have
$$
A^* A =
begin{pmatrix}
a_{1,1}^* & ldots & a_{n,1}^* \
vdots & ddots & vdots \
a_{1,n}^* & ldots & a_{n,n}^*
end{pmatrix}
begin{pmatrix}
a_{1,1} & ldots & a_{1,n} \
vdots & ddots & vdots \
a_{n,1} & ldots & a_{n,n}
end{pmatrix}
= begin{pmatrix}
sum_{i = 1}^{n} a_{i,1} a_{i,1}^* & & ast \
& ddots & \
ast & & sum_{i = 1}^{n} a_{i,n} a_{i,n}^*
end{pmatrix}
$$
therefore we have $operatorname{tr}(A^* A) = sum_{i,j=1}^{n} |a_{i,j}|^2$
- Let let $operatorname{tr}(A^*A) = 0$ then $|a_{i,j}|=0$ for all so the $A$ is the zero matrix, so $det(A)=0$, so it's not invertible, proving (1) by contraposition.
- From the above sum formula for the trace, since all summands are positive, when the sum is smaller than $n^2$, which is the number of summands, one of the summands has to be smaller than 1.
- Proven in (1).
answered yesterday
Viktor Glombik
474220
add a comment |
up vote
2
down vote
accepted
Let $A in mathbb{C}^{n times n}$ be a complex square matrix.
Since for a complex number $z = x + iy$ we have
$$
z^* z = (x - iy)(x + iy) = x^2 + y^2 = |z|^2,
quad text{where } |z| := sqrt{x^2 + y^2}
$$
Now we have
$$
A^* A =
begin{pmatrix}
a_{1,1}^* & ldots & a_{n,1}^* \
vdots & ddots & vdots \
a_{1,n}^* & ldots & a_{n,n}^*
end{pmatrix}
begin{pmatrix}
a_{1,1} & ldots & a_{1,n} \
vdots & ddots & vdots \
a_{n,1} & ldots & a_{n,n}
end{pmatrix}
= begin{pmatrix}
sum_{i = 1}^{n} a_{i,1} a_{i,1}^* & & ast \
& ddots & \
ast & & sum_{i = 1}^{n} a_{i,n} a_{i,n}^*
end{pmatrix}
$$
therefore we have $operatorname{tr}(A^* A) = sum_{i,j=1}^{n} |a_{i,j}|^2$
- Let let $operatorname{tr}(A^*A) = 0$ then $|a_{i,j}|=0$ for all so the $A$ is the zero matrix, so $det(A)=0$, so it's not invertible, proving (1) by contraposition.
- From the above sum formula for the trace, since all summands are positive, when the sum is smaller than $n^2$, which is the number of summands, one of the summands has to be smaller than 1.
- Proven in (1).
answered yesterday
Viktor Glombik
474220
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $A in mathbb{C}^{n times n}$ be a complex square matrix.
Since for a complex number $z = x + iy$ we have
$$
z^* z = (x - iy)(x + iy) = x^2 + y^2 = |z|^2,
quad text{where } |z| := sqrt{x^2 + y^2}
$$
Now we have
$$
A^* A =
begin{pmatrix}
a_{1,1}^* & ldots & a_{n,1}^* \
vdots & ddots & vdots \
a_{1,n}^* & ldots & a_{n,n}^*
end{pmatrix}
begin{pmatrix}
a_{1,1} & ldots & a_{1,n} \
vdots & ddots & vdots \
a_{n,1} & ldots & a_{n,n}
end{pmatrix}
= begin{pmatrix}
sum_{i = 1}^{n} a_{i,1} a_{i,1}^* & & ast \
& ddots & \
ast & & sum_{i = 1}^{n} a_{i,n} a_{i,n}^*
end{pmatrix}
$$
therefore we have $operatorname{tr}(A^* A) = sum_{i,j=1}^{n} |a_{i,j}|^2$
- Let let $operatorname{tr}(A^*A) = 0$ then $|a_{i,j}|=0$ for all so the $A$ is the zero matrix, so $det(A)=0$, so it's not invertible, proving (1) by contraposition.
- From the above sum formula for the trace, since all summands are positive, when the sum is smaller than $n^2$, which is the number of summands, one of the summands has to be smaller than 1.
- Proven in (1).
answered yesterday
Viktor Glombik
474220
Let $A in mathbb{C}^{n times n}$ be a complex square matrix.
Since for a complex number $z = x + iy$ we have
$$
z^* z = (x - iy)(x + iy) = x^2 + y^2 = |z|^2,
quad text{where } |z| := sqrt{x^2 + y^2}
$$
Now we have
$$
A^* A =
begin{pmatrix}
a_{1,1}^* & ldots & a_{n,1}^* \
vdots & ddots & vdots \
a_{1,n}^* & ldots & a_{n,n}^*
end{pmatrix}
begin{pmatrix}
a_{1,1} & ldots & a_{1,n} \
vdots & ddots & vdots \
a_{n,1} & ldots & a_{n,n}
end{pmatrix}
= begin{pmatrix}
sum_{i = 1}^{n} a_{i,1} a_{i,1}^* & & ast \
& ddots & \
ast & & sum_{i = 1}^{n} a_{i,n} a_{i,n}^*
end{pmatrix}
$$
therefore we have $operatorname{tr}(A^* A) = sum_{i,j=1}^{n} |a_{i,j}|^2$
- Let let $operatorname{tr}(A^*A) = 0$ then $|a_{i,j}|=0$ for all so the $A$ is the zero matrix, so $det(A)=0$, so it's not invertible, proving (1) by contraposition.
- From the above sum formula for the trace, since all summands are positive, when the sum is smaller than $n^2$, which is the number of summands, one of the summands has to be smaller than 1.
- Proven in (1).
answered yesterday
Viktor Glombik
474220
edited yesterday
answered yesterday
Viktor Glombik
474220
answered yesterday
Viktor Glombik
474220
answered yesterday
Viktor Glombik
474220
474220
add a comment |
add a comment |
math math is a new contributor. Be nice, and check out our Code of Conduct.
math math is a new contributor. Be nice, and check out our Code of Conduct.
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What is $A^*$ in context? Complex conjugate of $A$? Conjugate-transpose of $A$?
– Richard Martin
yesterday
@RichardMartin Conjugate transpose
– Jean-Claude Arbaut
yesterday
3
Hint: $mathrm{tr}(A^*A)=sum_{i,j} |a_{ij}|^2$. @RichardMartin No, it's not the "sum of squares", there is a modulus.
– Jean-Claude Arbaut
yesterday