# If A is an n squared matrix then show that $AA'$ and $A'A$ are Symmetric

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Hint-Use matrix properties

Any matrix is said to be symmetric if and only if:

$ \to $The matrix is a square matrix and

$ \to $The transpose of the matrix must be equal to itself.

Then here we know that the given Matrix A is a square matrix then the transpose of is A i.e. $A'$is also a square matrix. Here we know that $A'$ is our transpose matrix.

Proof:

$(AA')' = (A')'(A)$ [By using reversible law]

$(AA'$$)'$$ = AA'$ $[\because (A')' = A]$

$(AA'$$)'$$ = AA'$

By using matrix properties we can say that $ = AA'$ is symmetric

Similarly if $AA'$ is symmetric then $A'A$is also symmetric

Hence we proved that for any n squared matrix $A'A$ and $AA'$ are symmetric

NOTE: This problem can also be solved directly by stating the matrix properties as they have already proved, for which statement is enough.

Any matrix is said to be symmetric if and only if:

$ \to $The matrix is a square matrix and

$ \to $The transpose of the matrix must be equal to itself.

Then here we know that the given Matrix A is a square matrix then the transpose of is A i.e. $A'$is also a square matrix. Here we know that $A'$ is our transpose matrix.

Proof:

$(AA')' = (A')'(A)$ [By using reversible law]

$(AA'$$)'$$ = AA'$ $[\because (A')' = A]$

$(AA'$$)'$$ = AA'$

By using matrix properties we can say that $ = AA'$ is symmetric

Similarly if $AA'$ is symmetric then $A'A$is also symmetric

Hence we proved that for any n squared matrix $A'A$ and $AA'$ are symmetric

NOTE: This problem can also be solved directly by stating the matrix properties as they have already proved, for which statement is enough.

Last updated date: 22nd Sep 2023

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