1) Consider the $n \times n$ matrix $A=(a_{ij})$ with
$$
a_{ij}=
\begin{cases}
1,&\text{if } i 2) Suppose $A$ is an $n \times n$ matrix such that $A^2=A$. Show that $$rank(A)+rank(I-A)=n$$ where $I$ represents unit matrix of order $n$.
Two questions related to matrices
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matrices
vector-spaces
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1What have you tried? Is this homework? (Bonus: do these questions sound familiar?) – 2011-05-04
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0What facts you know about matrices and vector spaces??? – 2011-05-04
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2For 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 $v=Av+(I-A)v$. – 2011-05-04
1 Answers
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A neat argument for 2:
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 $ker(A)=im(A-I)$.