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

[nhrock3]

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

<br /> <br /> A*A=\sum_{k=1}^{n}a_{ik}a_{kj}<br /> <br />



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

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

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?

 

[micromass]

Hi nhrock3!

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

then A=0

This is false. Consider

A=\left(\begin{array}{cc} 0 &amp; 1\\ 0 &amp; 0\\ \end{array}\right)

then tr(A²)=0, but A is not zero.

What does the exercise say precisely?

 

[nhrock3]

if A is a simetric matrix and if tr(A^2)=0 then A=0

 

[micromass]

Ah, you should have said they were symmetric!

If

A=\left(\begin{array}{ccccc}<br /> a_{11} &amp; a_{12} &amp; a_{13} &amp; ... &amp; a_{1n}\\<br /> a_{12} &amp; a {22} &amp; a_{23} &amp; ... &amp; a_{2n}\\<br /> a_{13} &amp; a_{23} &amp; a_{33} &amp; ... &amp; a_{3n}\\<br /> \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots\\<br /> a_{1n} &amp; a_{2n} &amp; a_{3n} &amp; ... &amp; a_{nn}\\<br /> \end{array}\right)

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

 

[nhrock3]

Ah, you should have said they were symmetric!

If

A=\left(\begin{array}{ccccc}<br /> a_{11} &amp; a_{12} &amp; a_{13} &amp; ... &amp; a_{1n}\\<br /> a_{12} &amp; a {22} &amp; a_{23} &amp; ... &amp; a_{2n}\\<br /> a_{13} &amp; a_{23} &amp; a_{33} &amp; ... &amp; a_{3n}\\<br /> \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots\\<br /> a_{1n} &amp; a_{2n} &amp; a_{3n} &amp; ... &amp; a_{nn}\\<br /> \end{array}\right)

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
tr(A^{2})=\sum_{k=1}^{n}\sum_{i=1}^{n}a_{ki}a_{ik}
so because its simetric and equlas zero
tr(A^{2})=\sum_{k=1}^{n}\sum_{i=1}^{n}(a_{ki})^2=0
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 :)

 

[micromass]

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

\|(a_{1k},...,a_{nk})\|^2

Do you see that?