$A$ is similar to $Diag(a_1,…,a_n)$. Then the sum of elements in A's diagonal is $a_1+…+a_n$
$begingroup$
Let $D=Diag(a_1,...,a_n)$ denote the diagonal matrix with diagonal $a_1,a_2,...,a_n$. I noticed that if $A$ and $D$ are similar marices, then the sum of elements in $A$'s diagonal is also $a_1+...+a_n$.
I could not find a counter example, and I'm not sure how to prove it. Is it always true? If so, I would love to see a proof using only the fact that there exists $P$ such that $A=P^{-1} cdot D cdot P$,if there is one, because that's all I know about similar matrices right now. Thank you!
linear-algebra matrices
asked Dec 29 '18 at 17:45
OmerOmer
3619
$endgroup$
add a comment |
$begingroup$
Let $D=Diag(a_1,...,a_n)$ denote the diagonal matrix with diagonal $a_1,a_2,...,a_n$. I noticed that if $A$ and $D$ are similar marices, then the sum of elements in $A$'s diagonal is also $a_1+...+a_n$.
I could not find a counter example, and I'm not sure how to prove it. Is it always true? If so, I would love to see a proof using only the fact that there exists $P$ such that $A=P^{-1} cdot D cdot P$,if there is one, because that's all I know about similar matrices right now. Thank you!
linear-algebra matrices
asked Dec 29 '18 at 17:45
OmerOmer
3619
$endgroup$
1
$begingroup$
The sum of the diagonal is called trace. It satisfies $tr(AB)=tr(BA)$ , just calculate. Therefore the trace of similar matrices are equal.
$endgroup$
– Card_Trick
Dec 29 '18 at 17:49
add a comment |
$begingroup$
Let $D=Diag(a_1,...,a_n)$ denote the diagonal matrix with diagonal $a_1,a_2,...,a_n$. I noticed that if $A$ and $D$ are similar marices, then the sum of elements in $A$'s diagonal is also $a_1+...+a_n$.
I could not find a counter example, and I'm not sure how to prove it. Is it always true? If so, I would love to see a proof using only the fact that there exists $P$ such that $A=P^{-1} cdot D cdot P$,if there is one, because that's all I know about similar matrices right now. Thank you!
linear-algebra matrices
asked Dec 29 '18 at 17:45
OmerOmer
3619
$endgroup$
Let $D=Diag(a_1,...,a_n)$ denote the diagonal matrix with diagonal $a_1,a_2,...,a_n$. I noticed that if $A$ and $D$ are similar marices, then the sum of elements in $A$'s diagonal is also $a_1+...+a_n$.
I could not find a counter example, and I'm not sure how to prove it. Is it always true? If so, I would love to see a proof using only the fact that there exists $P$ such that $A=P^{-1} cdot D cdot P$,if there is one, because that's all I know about similar matrices right now. Thank you!
linear-algebra matrices
linear-algebra matrices
asked Dec 29 '18 at 17:45
OmerOmer
3619
asked Dec 29 '18 at 17:45
OmerOmer
3619
asked Dec 29 '18 at 17:45
OmerOmer
3619
asked Dec 29 '18 at 17:45
OmerOmer
3619
asked Dec 29 '18 at 17:45
OmerOmer
3619
3619
1
$begingroup$
The sum of the diagonal is called trace. It satisfies $tr(AB)=tr(BA)$ , just calculate. Therefore the trace of similar matrices are equal.
$endgroup$
– Card_Trick
Dec 29 '18 at 17:49
add a comment |
1
$begingroup$
The sum of the diagonal is called trace. It satisfies $tr(AB)=tr(BA)$ , just calculate. Therefore the trace of similar matrices are equal.
$endgroup$
– Card_Trick
Dec 29 '18 at 17:49
1
1
$begingroup$
The sum of the diagonal is called trace. It satisfies $tr(AB)=tr(BA)$ , just calculate. Therefore the trace of similar matrices are equal.
$endgroup$
– Card_Trick
Dec 29 '18 at 17:49
$begingroup$
The sum of the diagonal is called trace. It satisfies $tr(AB)=tr(BA)$ , just calculate. Therefore the trace of similar matrices are equal.
$endgroup$
– Card_Trick
Dec 29 '18 at 17:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The sum of the diagonal elements of a matrix is called the trace of a matrix. Use the decomposition $A = P^{-1} D P$ as well as the following property of trace:
$$text{tr}(M_1 M_2 M_3) = text{tr}(M_2 M_3 M_1).$$
answered Dec 29 '18 at 17:47
angryavianangryavian
41.6k23381
$endgroup$
3
$begingroup$
To prove the mentioned property you only need the simpler $text{tr}(AB) = text{tr}(BA)$.
$endgroup$
– Michh
Dec 29 '18 at 17:49
$begingroup$
@angryavian Nice, thanks. The proof of that property is just as technical as multiply those matrices and see the diagonal sum is the same?
$endgroup$
– Omer
Dec 29 '18 at 17:49
1
$begingroup$
@Omer Yes, you can prove it directly in that manner. And as others mentioned, it suffices to just deal with the product of two matrices.
$endgroup$
– angryavian
Dec 29 '18 at 17:56
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The sum of the diagonal elements of a matrix is called the trace of a matrix. Use the decomposition $A = P^{-1} D P$ as well as the following property of trace:
$$text{tr}(M_1 M_2 M_3) = text{tr}(M_2 M_3 M_1).$$
answered Dec 29 '18 at 17:47
angryavianangryavian
41.6k23381
$endgroup$
3
$begingroup$
To prove the mentioned property you only need the simpler $text{tr}(AB) = text{tr}(BA)$.
$endgroup$
– Michh
Dec 29 '18 at 17:49
$begingroup$
@angryavian Nice, thanks. The proof of that property is just as technical as multiply those matrices and see the diagonal sum is the same?
$endgroup$
– Omer
Dec 29 '18 at 17:49
1
$begingroup$
@Omer Yes, you can prove it directly in that manner. And as others mentioned, it suffices to just deal with the product of two matrices.
$endgroup$
– angryavian
Dec 29 '18 at 17:56
add a comment |
$begingroup$
The sum of the diagonal elements of a matrix is called the trace of a matrix. Use the decomposition $A = P^{-1} D P$ as well as the following property of trace:
$$text{tr}(M_1 M_2 M_3) = text{tr}(M_2 M_3 M_1).$$
answered Dec 29 '18 at 17:47
angryavianangryavian
41.6k23381
$endgroup$
3
$begingroup$
To prove the mentioned property you only need the simpler $text{tr}(AB) = text{tr}(BA)$.
$endgroup$
– Michh
Dec 29 '18 at 17:49
$begingroup$
@angryavian Nice, thanks. The proof of that property is just as technical as multiply those matrices and see the diagonal sum is the same?
$endgroup$
– Omer
Dec 29 '18 at 17:49
1
$begingroup$
@Omer Yes, you can prove it directly in that manner. And as others mentioned, it suffices to just deal with the product of two matrices.
$endgroup$
– angryavian
Dec 29 '18 at 17:56
add a comment |
$begingroup$
The sum of the diagonal elements of a matrix is called the trace of a matrix. Use the decomposition $A = P^{-1} D P$ as well as the following property of trace:
$$text{tr}(M_1 M_2 M_3) = text{tr}(M_2 M_3 M_1).$$
answered Dec 29 '18 at 17:47
angryavianangryavian
41.6k23381
$endgroup$
The sum of the diagonal elements of a matrix is called the trace of a matrix. Use the decomposition $A = P^{-1} D P$ as well as the following property of trace:
$$text{tr}(M_1 M_2 M_3) = text{tr}(M_2 M_3 M_1).$$
answered Dec 29 '18 at 17:47
angryavianangryavian
41.6k23381
answered Dec 29 '18 at 17:47
angryavianangryavian
41.6k23381
answered Dec 29 '18 at 17:47
angryavianangryavian
41.6k23381
answered Dec 29 '18 at 17:47
angryavianangryavian
41.6k23381
41.6k23381
3
$begingroup$
To prove the mentioned property you only need the simpler $text{tr}(AB) = text{tr}(BA)$.
$endgroup$
– Michh
Dec 29 '18 at 17:49
$begingroup$
@angryavian Nice, thanks. The proof of that property is just as technical as multiply those matrices and see the diagonal sum is the same?
$endgroup$
– Omer
Dec 29 '18 at 17:49
1
$begingroup$
@Omer Yes, you can prove it directly in that manner. And as others mentioned, it suffices to just deal with the product of two matrices.
$endgroup$
– angryavian
Dec 29 '18 at 17:56
add a comment |
3
$begingroup$
To prove the mentioned property you only need the simpler $text{tr}(AB) = text{tr}(BA)$.
$endgroup$
– Michh
Dec 29 '18 at 17:49
$begingroup$
@angryavian Nice, thanks. The proof of that property is just as technical as multiply those matrices and see the diagonal sum is the same?
$endgroup$
– Omer
Dec 29 '18 at 17:49
1
$begingroup$
@Omer Yes, you can prove it directly in that manner. And as others mentioned, it suffices to just deal with the product of two matrices.
$endgroup$
– angryavian
Dec 29 '18 at 17:56
3
3
$begingroup$
To prove the mentioned property you only need the simpler $text{tr}(AB) = text{tr}(BA)$.
$endgroup$
– Michh
Dec 29 '18 at 17:49
$begingroup$
To prove the mentioned property you only need the simpler $text{tr}(AB) = text{tr}(BA)$.
$endgroup$
– Michh
Dec 29 '18 at 17:49
$begingroup$
@angryavian Nice, thanks. The proof of that property is just as technical as multiply those matrices and see the diagonal sum is the same?
$endgroup$
– Omer
Dec 29 '18 at 17:49
$begingroup$
@angryavian Nice, thanks. The proof of that property is just as technical as multiply those matrices and see the diagonal sum is the same?
$endgroup$
– Omer
Dec 29 '18 at 17:49
1
1
$begingroup$
@Omer Yes, you can prove it directly in that manner. And as others mentioned, it suffices to just deal with the product of two matrices.
$endgroup$
– angryavian
Dec 29 '18 at 17:56
$begingroup$
@Omer Yes, you can prove it directly in that manner. And as others mentioned, it suffices to just deal with the product of two matrices.
$endgroup$
– angryavian
Dec 29 '18 at 17:56
add a comment |
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The sum of the diagonal is called trace. It satisfies $tr(AB)=tr(BA)$ , just calculate. Therefore the trace of similar matrices are equal.
$endgroup$
– Card_Trick
Dec 29 '18 at 17:49