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Definitions: For $F$ a field,

  • $A,B\in F^{m\times n}$ are equivalent means there exist invertible matrices $Q\in F^{m\times m}$ and $P\in F^{n\times n}$ so that $B=QAP$,
  • $A,B\in F^{n\times n}$ are similar means there exists an invertible matrix $P\in F^{n\times n}$ so that $B=PAP^{-1}$.
  • $[T]_\mathcal{B}$ is the matrix of the linear transformation $T:V\to V$ with respect to an ordered basis $\mathcal B$ of $V$.

I've seen the result that $A,B\in F^{m\times n}$ are equivalent iff they have the same row and column reduced echelon form iff they have the same rank.

I'd like to show that if $A,B\in F^{n\times n}$ are idempotent matrices, they are similar iff they have the same rank.

I've noted that similarity of two matrices implies their equivalence, and hence that they have the same rank. So it remains to show that $rank(A)=rank(B)$ implies $A$ and $B$ are similar.

Observations: If $V$ is a finite-dimensional vector space over field $F$ and $T:V\to V$ is an idempotent linear transformation (i.e. $T^2=T$), then $V=im T\oplus \ker T$ and if $\mathcal B_1=\{v_1,v_2,...,v_m\},\mathcal B_2=\{v_{m+1},...,v_n\}$ are bases for $im T$ and $\ker T$ respectively, then $[T]_{\mathcal B_1\cup \mathcal B_2}\in F^{n\times n}$ is a $1\times 2$ block matrix with $C\in F^{n\times m}$ in the first entry and zero matrix for the other entry, where the $j$th column of $C$ is given by the scalars $a_{1j},...,a_{nj}$ so that $T(v_j)=\sum_{i=1}^n a_{ij}v_i$ (which uniquely define $T(v_j)$ since $\{v_1,...,v_n\}$ are a basis for $V$).

Let $L_A,L_B$ be the linear transformations representing left-multiplication by $A$ and $B$, respectively. Then $A'=[L_A]_\mathcal B$ and $B'=[L_B]_\mathcal B$ take the block matrix form described above, and since $rank (A)=rank (B)$ by assumption, $A'$ and $B'$ have the same row and column reduced echelon form. So starting from the row and column reduced echelon form that they take, we can multiply on the left and right by invertible matrices (by doing the inverse elementary matrix operations that we did to get the reduced forms) to get back to either matrix, i.e. if $D$ is the row and column reduced echelon form for the two, then $QA'P=D=RB'S$ where $Q,P,R,S$ are invertible $n\times n$ matrices.

Hence $B'=R^{-1}QA'PS^{-1}$. We want to show that the $A,B$ being idempotent gives us that $R^{-1}Q=(PS^{-1})^{-1}$. This is where I'm stuck.

Any help would be greatly appreciated! Thanks in advance

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  • $\begingroup$ As a hint rather than a full answer, note that the result would imply that any idempotent $T:V \to V$ is given by $T = \operatorname{diag}(1, \dots, 1, 0, \dots, 0)$ over some basis. Try to find that basis explicitly. $\endgroup$
    – anomaly
    Commented Nov 6, 2016 at 2:17

5 Answers 5

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If $P$ is idempotent with rank $r$ then there is a basis $b_k$ such that $P b_k = b_k$ for $k=1,...,r$ and $P b_k = 0$ otherwise.

Suppose $A,B$ are idempotent with the same rank $r$, then there are bases $\alpha_k, \beta_k$ such that the above relations hold.

Define $Q\alpha_k = \beta _k$, then for $k \le r$ we have $QA \alpha_k = Q \alpha_k = \beta_k = B \beta_k = BQ \alpha_k$. For $r >k$ we have $Q A \alpha_k = 0 = B \beta _k = B Q \alpha_k$. Note that $Q$ is invertible.

In particular we have $QA = BQ$ and so $A,B$ are similar.

For the other direction, if $A,B$ are similar, then they have the same rank (this is true for any matrices $A,N$, not just idempotent ones). Suppose there is some $Q$ such that $QA = BQ$ and suppose $b_1,...,b_r$ is a basis for the range of $B$. Then, since $A = Q^{-1} BQ$, we see that $ Q^{-1} b_1,...,Q^{-1} b_r$ is a basis for the range of $A$. Hence both have the same rank.

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  • $\begingroup$ Excellent, thanks a bunch! Could you please clarify how we get the existence and invertibility of $Q$? $\endgroup$
    – manofbear
    Commented Nov 5, 2016 at 23:02
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    $\begingroup$ I have shown the existence above, to define a linear operator, it is sufficient to define it on a basis. I have defined it on $\alpha_k$ above. If $Ax = 0$ then $x = \sum x_i (A\alpha_k)= \sum x_i (\beta_k) = 0$ and since the $\beta_k$ are a basis, it follows that $x=0$. $\endgroup$
    – copper.hat
    Commented Nov 5, 2016 at 23:06
  • $\begingroup$ Sorry, I'm still missing it. This shows $A$ is invertible, right? But we want $Q$ invertible. $A$ invertible doesn't imply $QA$ invertible, so there must be something I'm not seeing. Could you point it out? $\endgroup$
    – manofbear
    Commented Nov 6, 2016 at 19:56
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    $\begingroup$ My mistake, replace A by Q in my previous comment above. $\endgroup$
    – copper.hat
    Commented Nov 6, 2016 at 20:28
  • $\begingroup$ Gotcha, figured out that's what we wanted literally within a minute of your response haha. Thanks a bunch! This has really clarified similar matrices for me, something that has been frustrating me the last few days $\endgroup$
    – manofbear
    Commented Nov 6, 2016 at 20:29
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Let $T:V \to V$ be an idempotent map, and let $V_0 = \ker T$ and $V_1 = T(V)$. Then $T\vert V_0 = 0$ and $T\vert V_1 = 1$, so $V = V_0 \oplus V_1$; furthermore, over a suitable basis, $T = 1_{r\times r} \oplus 0$, where $r = \dim V_1 = \operatorname{rank} T$.

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Here is another proof by showing that each idempotent matrix $A$ is similar to a diagonal matrix, i.e $$ P^{-1}AP=\pmatrix{I_r &0\\0 &0}\tag1 $$ where $r$ is the multiplicity of eigenvalue $1$ of $A$.

If $A$ is idempotent, then $A^2-A=0$. So $m_A|x^2-x$, where $m_A(x)$ is the minimum polynomial of $A$. Thus the eigenvalues of $A$ are $0$ and $1$. Since $x^2-x=x(x-1)$ is of linear order, $A$ is similar to a diagonal matrix. So $(1)$ follows with $r$ being the number of eigenvalue $1$ of $A$.

Clearly, $\operatorname{Rank}(A)=r$. Thus by $(1)$, idempotent matrices with the same rank are similar to the same matrix and are similar.

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Similar matrices have the same rank, so one implication is done with.

Suppose $P$ is idempotent. Then, denoting by $C(P)$ the column space and by $N(P)$ the null space, we easily see that $F^n=C(P)\oplus N(P)$. Indeed any element $v\in F^n$ (vectors are represented as columns) can be written as $$ v=Pv+(v-Pv) $$ and $P(v-Pv)=0$; so $F^n=C(P)+N(P)$. Moreover, if $v=Pw\in C(P)$ and $Pv=0$, we have $v=Pw=P^2w=Pv=0$. Thus $C(P)\cap N(P)=\{0\}$.

Note that every vector in $C(P)$ is an eigenvector for $P$ with respect to the eigenvalue $1$; any vector in $N(P)$ is an eigenvector of $P$ with respect to the eigenvalue $0$. Since the geometric multiplicity of an eigenvalue is at most equal to the algebraic multiplicity, we conclude that $P$ is diagonalizable and it has the eigenvalue $1$ with multiplicity $k$ (the rank of $P$) and the eigenvalue $0$ with multiplicity $n-k$.

Matrices similar to the same matrix are similar.

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A more linear-algebraic approach.

If $A^2=A$ and $B^2=B$ for linear transformations $A,B:V\to V$, with $V$ and $n$-dimensional vector space, then $\mathrm{rank}(A)=\mathrm{rank}(B)$ means that the dimensions of $AV$ and $BV$ are the same. This means that there is an isomorphism:

$$\phi:AV\to BV$$

Now, let $\ker(A)=\{v\mid Av=0\}$, and $\ker(B)=\{v\mid Bv=0\}$.

Claim: $AV\cap\ker A=\{0\}$.

If $v\in AV$, then $v=Aw$ for some $w$. If $v\in\ker A$, then $0=Av=A^2w=Aw=v$. So $v=0$ if $v\in AV\cap\ker A$.

Claim: $V=AV + \ker A$

If $v\in V$ then $v=Av + (v-Av)$ and $A(v-Av)=Av-A^2v=0$ so $v-Av\in\ker A$.

This means $\dim(\ker A)=n-\dim(AV)=n-\dim(BV)=\dim(\ker B)$.

So there is an isomorphism:

$$\rho: \ker A\to \ker B$$

Now, define $C:V\to V$ as:

$$Cv = \phi(Av) + \rho(v-Av)$$

Show that $C$ Is invertible, with:

$$C^{-1}v = \phi^{-1}(Bv) +\rho^{-1}(v-Bv)$$

Finally, show that $C^{-1}BC=A$.

This last is because $BCv= B\phi(Av)+B\rho(v-Av)=B\phi(Av)=\phi(Av)$, since $\rho(v-Av)\in \ker B$, and $\phi(Av)\in BV$ means $B\phi(Av)=\phi(Av)$.

Applying $C^{-1}$ to this gives us $C^{-1}BCv=\phi^{-1}(\phi(Av))=Av$.

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