Proving that A=0 When tr(A^2)=0

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In summary, the conversation discusses how to prove that if the trace of a symmetric matrix A squared is equal to zero, then A must be equal to zero. It is mentioned that when multiplying two symmetric matrices, the resulting matrix has a diagonal where each element is the dot product of the corresponding row and column. The formula for the trace of A squared is also discussed and it is shown that because the matrix is symmetric, the trace is equal to the sum of squares of the elements on the diagonal. Finally, it is concluded that in order for the sum of squares to equal zero, each individual element on the diagonal must be equal to zero.
  • #1
nhrock3
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i need to prove that if tr(A^2)=0

then A=0



we have a multiplication of 2 the same simmetrical matrices

why there multiplication is this sum formula

[iTEX]

A*A=\sum_{k=1}^{n}a_{ik}a_{kj}

[/iTEX]



i know that wjen we multiply two matrices then in our result matrix

each aij member is dot product of i row and j column.

dont understand the above formula.





and i don't understand how they got the following formula:

then when we calculate the trace (the sum of the diagonal members)

we get

[TEX]

tr(A^{2})=\sum_{i=1}^{n}A_{ii}^{2}=\sum_{i=1}^{n}(\sum_{i=1}^{n}a_{ik}a_{ki})

[/TEX]

and because the matrix is simmetric then the trace is zero

why?


i need to prove that if tr(A^2)=0



then A=0



can you explain the sigma work in order to prove it?
 
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  • #2
Hi nhrock3! :smile:

nhrock3 said:
i need to prove that if tr(A^2)=0

then A=0

This is false. Consider

[tex]A=\left(\begin{array}{cc} 0 & 1\\ 0 & 0\\ \end{array}\right)[/tex]

then tr(A2)=0, but A is not zero.

What does the exercise say precisely?
 
  • #3
if A is a simetric matrix and if tr(A^2)=0 then A=0
 
  • #4
Ah, you should have said they were symmetric! :smile:

If

[tex]A=\left(\begin{array}{ccccc}
a_{11} & a_{12} & a_{13} & ... & a_{1n}\\
a_{12} & a {22} & a_{23} & ... & a_{2n}\\
a_{13} & a_{23} & a_{33} & ... & a_{3n}\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
a_{1n} & a_{2n} & a_{3n} & ... & a_{nn}\\
\end{array}\right)[/tex]

then what will be the diagonal of A2?? In particular, can you show that the diagonal contains only positive values?
 
  • #5
micromass said:
Ah, you should have said they were symmetric! :smile:

If

[tex]A=\left(\begin{array}{ccccc}
a_{11} & a_{12} & a_{13} & ... & a_{1n}\\
a_{12} & a {22} & a_{23} & ... & a_{2n}\\
a_{13} & a_{23} & a_{33} & ... & a_{3n}\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
a_{1n} & a_{2n} & a_{3n} & ... & a_{nn}\\
\end{array}\right)[/tex]

then what will be the diagonal of A2?? In particular, can you show that the diagonal contains only positive values?

each member of the diagonal on the A^2 matrix its number of row equals the column number

so the second member in the diagonal is the dot product of the second row with the second column
etc..
so i get this expression
[tex]tr(A^{2})=\sum_{k=1}^{n}\sum_{i=1}^{n}a_{ki}a_{ik}[/tex]
so because its simetric and equlas zero
[tex]tr(A^{2})=\sum_{k=1}^{n}\sum_{i=1}^{n}(a_{ki})^2=0[/tex]
so we get a sum of squeres and in order for them to be zero
then each one of them has to be zero
thannkkks :)
 
Last edited:
  • #6
nhrock3 said:
each member of the diagonal on the A^2 matrix its number of row equals the column number

so the second member in the diagonal is the dot product of the second row with the second column

Indeed, so the k'th member in the diagonal is

[tex]\|(a_{1k},...,a_{nk})\|^2[/tex]

Do you see that?
 

FAQ: Proving that A=0 When tr(A^2)=0

How can I prove that A=0 when tr(A^2)=0?

To prove that A=0 when tr(A^2)=0, you can use the fact that the trace of a matrix is equal to the sum of its eigenvalues. Since the trace of A^2 is equal to 0, the sum of the eigenvalues of A^2 must also be 0. This means that all the eigenvalues of A^2 are equal to 0. Since the eigenvalues of A^2 are the squares of the eigenvalues of A, this means that all the eigenvalues of A are also equal to 0. And when all the eigenvalues of a matrix are 0, the matrix itself must be the zero matrix.

Can you provide an example to illustrate this proof?

Sure, let's take the matrix A = [0 0; 0 0]. The trace of A^2 is equal to 0+0=0. The eigenvalues of A are both equal to 0, and the eigenvalues of A^2 are both equal to 0^2=0. This example shows that when tr(A^2)=0, A must be equal to the zero matrix.

What is the significance of the trace of a matrix in this proof?

The trace of a matrix is important because it represents the sum of the diagonal elements of the matrix. In this proof, we use the fact that the trace is equal to the sum of the eigenvalues of a matrix, which helps us to understand the behavior of the eigenvalues in relation to the trace.

Can this proof be applied to matrices of any size?

Yes, this proof can be applied to matrices of any size. The fact that the trace of a matrix is equal to the sum of its eigenvalues holds for all square matrices, regardless of their dimension.

Are there any other conditions in which A=0 when tr(A^2)=0?

Yes, there is another condition in which A=0 when tr(A^2)=0. If A is a symmetric matrix, then tr(A^2)=0 also implies that A=0. This is because the eigenvalues of a symmetric matrix are always real, and the only real number that squares to 0 is 0 itself. Therefore, if tr(A^2)=0 and A is symmetric, then all the eigenvalues of A must be 0, making A the zero matrix.

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