Idempotent Matrix

**Xingyuan** · October 19th 2009, 05:54 PM

If $A$ is a $n \times n$ matrix, and satisfy $A^{2} = A$ .

show that: $rank(A) = tr(A)$

vice verse, If matrix $A$ satisfy the condition : $rank(A) = tr(A)$ , $(*)$
then $A$ not always Idempotent matrix, such as: $\left(\begin{array}{cc}1 & 2 \\<br /> 1 & 1\end{array}\right)$ (corrected)
can we add some conditions to $"(*)"$ such that $A$ is a Idempotent matrix

**NonCommAlg** · October 19th 2009, 07:58 PM

Originally Posted by Xingyuan

If $A$ is a $n \times n$ matrix, and satisfy $A^{2} = A$ .

show that: $rank(A) = tr(A)$

i don't think this is true for all idempotent matrices. my guess is that we also need $A$ to be symmetric.

**Noxide** · October 19th 2009, 08:01 PM

what does tr(A) mean? lol...

**Gamma** · October 19th 2009, 08:52 PM

Tr(A) is the sum of the elements along the diagonal.

**tonio** · October 19th 2009, 10:54 PM

Originally Posted by Xingyuan

If $A$ is a $n \times n$ matrix, and satisfy $A^{2} = A$ .

show that: $rank(A) = tr(A)$

vice verse, If matrix $A$ satisfy the condition : $rank(A) = tr(A)$ , $(*)$
then $A$ not always Idempotent matrix, such as: $\left(\begin{array}{cc}1 & 1 \\<br /> 1 & 1\end{array}\right)$
can we add some conditions to $"(*)"$ such that $A$ is a Idempotent matrix

Well, if the matrix is $2x2$ the it is true since if $A=\left(\begin{array}{cc}a&b\\c&d\end{array}\right )$ , then $A^2=A\Longrightarrow\,\,a+d=1\,\,\mbox{or}\,\,b=c= 0\,,\,\,\mbox{or}\,\,\,a=d =\frac{1}{2}\,,\,\,bc=\frac{1}{4}$

And since from the condition it follows that the minimal pol. of $A$ divides $x(x-1)$ , we get what we want (if the matrix has rank 2 then zero can't be one of its eigenvectors thus its characteristic pol. is $(x-1)^2\Longrightarrow\,\mbox{from its Jordan Canonical form...etc.}$

For matrices of order greater than 2 I'm not sure but I doubt it.

Tonio

**Opalg** · October 20th 2009, 12:02 PM

Originally Posted by Xingyuan

If $A$ is a $n \times n$ matrix, and satisfy $A^{2} = A$ .

show that: $rank(A) = tr(A)$

Think of A as acting as a linear transformation on an n-dimensional vector space V. Then $\text{ran}(A)\cap\text{ker}(A) = (0)$ and $\text{ran}(A)+\text{ker}(A) = V$ . The reason for this is that if $y=Ax\in\text{ran}(A)\cap\text{ker}(A)$ then $y = Ax = A^2x = Ay=0$ . Also, if $z\in V$ then $z = Az + (z-Az) \in\text{ran}(A)+\text{ker}(A)$ .

Now let $\{e_1,\ldots,e_k\}$ be a basis for ran(A) (so that k = rank(A)), and let $\{e_{k+1},\ldots,e_n\}$ be a basis for ker(A). Then $\{e_1,\ldots,e_n\}$ is a basis for V, and the matrix of the linear transformation A with respect to this basis is a diagonal matrix with k 1s in the first k places on the diagonal, and 0s in the remaining places on the diagonal. This matrix obviously has trace k. But it is similar to the original matrix A, and so has the same trace as A. Therefore trace(A) = k = rank(A).

Originally Posted by Xingyuan

vice verse, If matrix $A$ satisfy the condition : $rank(A) = tr(A)$ , $(*)$
then $A$ not always Idempotent matrix, such as: $\left(\begin{array}{cc}1 & 2 \\<br /> 1 & 1\end{array}\right)$ (corrected)
can we add some conditions to $"(*)"$ such that $A$ is a Idempotent matrix

If B is an arbitrary n×n matrix with nonzero trace, then there is a scalar multiple of B, namely $(k/\text{tr}(B))B$ , that has trace k. So knowing that a matrix has trace k tells you practically nothing about the structure of the matrix, and it will be no help at all in determining whether or not B is an idempotent.

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