# Two questions related to matrices

1) Consider the n x n matrix A=((aij)) with aij=1 for i < j and aij=0 for i≥j.

Let V = {f(A): f is a polynomial with real coefficients}. Note that V is a vector space with usual operations. Find the dimension of V, when (a) n=3, and (b) n=4.

2) Suppose A is an n x n matrix such that A2=A. Show that rank(A)+rank(I-A)=n where I represents unit matrix of order n.

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What have you tried? Is this homework? (Bonus: do these questions sound familiar?) –  Did May 4 '11 at 14:30
What facts you know about matrices and vector spaces??? –  Fabian May 4 '11 at 14:42
For question 1: in the $n=3$ case, compute the first few powers of the matrix $A$. Does this suggest anything you might say about the higher powers of $A$? Now consider what happens when you substitute $A$ for $x$ in a polynomial with real coefficients. What will the degree of the polynomial in $A$ actually be? The answer should suggest what the dimension of $V$ is. The $n=4$ case is smilar. For the second question, you might note that any vector $v$ can be expressed as $v=Av+(I-A)v$. –  Chris Leary May 4 '11 at 14:58

What formula do we know involving rank and $n$? Well, the Rank-Nullity Theorem is close ($rank+null=n$).
So we know that $rank(A)+null(A)=n$. Hence, if $ker(A)=im(A-I)$, we would be done. Well, if $v=(A-I)w$, then it is clear that $Av=A(A-I)w=(A^2-A)w=0w=0$ so $im(A-I)\subset ker(A)$. In the other direction, note that if $Av=0$, then $v=(A-I)(-v)$ so $v\in im(A-I)$. Hence $ker(A)\subset im(A-I)$ and we conclude that $ker(A)=im(A-I)$.