Let A be a 3*3 matrix with A^3 = I. Find the rank of the matrix B = A - I.
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According to your class-notes/textbook - what tis the definition of the rank of a matrix?
Please show us what you have tried and exactly where you are stuck.
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The rank of a matrix is the maximum number of its linearly independent column vectors
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You have found the right formulas but it is hard to follow your thoughts. We have
[math] 0=A^3-I=(A-I)(A^2+A+I)=B(A^2+A+I)=(A^2+A+I)(A-I)=(A^2+A+I)B [/math]This means that [imath] B [/imath] sends the image of [imath] A^2+A+I [/imath] to [imath] \{0\} [/imath] and that the image of [imath] B [/imath] is sent to [imath] \{0\} [/imath] by [imath] A^2+A+I. [/imath]
We are interested in the dimension of the image of [imath] B. [/imath] This has to be [imath] 0,1,2 [/imath] or [imath] 3. [/imath] From the above we know that
[math] \begin{gathered} \operatorname{im}(A^2+A+I) \subseteq \operatorname{ker}(B)\\ \operatorname{im}(B)\subseteq \operatorname{ker}(A^2+A+I)\\ 3=\operatorname{im}(B)+\operatorname{ker}(B)\\ 3=\operatorname{im}(A^2+A+I)+\operatorname{ker}(A^2+A+I) \end{gathered} [/math]
This is what we get from what you wrote in your image. The next questions are:
What do we know about the eigenvalues of [imath] A\, , \,B\, , \,A^2+A+I=A^{-1}+A+I [/imath]?
Can we rule out some possibilities for [imath] \operatorname{rank}(B)=\dim \operatorname{im}(B) [/imath] among [imath] \{0,1,2,3\} [/imath] and why?
[math] 0=A^3-I=(A-I)(A^2+A+I)=B(A^2+A+I)=(A^2+A+I)(A-I)=(A^2+A+I)B [/math]This means that [imath] B [/imath] sends the image of [imath] A^2+A+I [/imath] to [imath] \{0\} [/imath] and that the image of [imath] B [/imath] is sent to [imath] \{0\} [/imath] by [imath] A^2+A+I. [/imath]
We are interested in the dimension of the image of [imath] B. [/imath] This has to be [imath] 0,1,2 [/imath] or [imath] 3. [/imath] From the above we know that
[math] \begin{gathered} \operatorname{im}(A^2+A+I) \subseteq \operatorname{ker}(B)\\ \operatorname{im}(B)\subseteq \operatorname{ker}(A^2+A+I)\\ 3=\operatorname{im}(B)+\operatorname{ker}(B)\\ 3=\operatorname{im}(A^2+A+I)+\operatorname{ker}(A^2+A+I) \end{gathered} [/math]
This is what we get from what you wrote in your image. The next questions are:
What do we know about the eigenvalues of [imath] A\, , \,B\, , \,A^2+A+I=A^{-1}+A+I [/imath]?
Can we rule out some possibilities for [imath] \operatorname{rank}(B)=\dim \operatorname{im}(B) [/imath] among [imath] \{0,1,2,3\} [/imath] and why?
Thank you very much for your reply!!
blamocur
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I don't see how this replacement helps us. I believe I solved the problem using eigenvalues and eigenvectors of [imath]A[/imath].
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blamocur
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Posting my solution one week after OP.
Claim: If [imath]A^3=I[/imath] then either [imath]A-I=0[/imath] or the rank of [imath]A-I[/imath] is 2.
Proof: Since [imath]A^3 = I[/imath] then for all 3 eigenvalues we have [imath]\lambda_i^3=1[/imath]. Because [imath]A[/imath] is real this means that either [imath]\forall i: \lambda_i = 1[/imath] or we have [imath]\lambda_0=1[/imath] and [imath]\lambda_1,\lambda_2[/imath] are two other (complex) cube roots of 1.
If [imath]\forall i:\lambda_i = 1[/imath] then [imath]\forall v \in R^3 : Av = v \Longrightarrow A=I[/imath], in which case [imath]A-I=0[/imath] has rank 0.
If all [imath]\lambda_i[/imath] are different then we'll show that the rank of [imath]A-I[/imath] is 2. Let [imath]g_1, g_2, g_3[/imath] be eigenvectors of [imath]A[/imath], and [imath]Ag_1 = g_1[/imath] or [imath](A-I)g_1 = 0[/imath]. This means that the rank is less than 3. We prove that the rank is 2 by contradiction: assume that it is less than 2, which means that is another vector [imath]v[/imath] such that [imath]v\neq \alpha v[/imath] for which [imath](A-I)v=0[/imath], or [imath]Av=v[/imath].
Since [imath]g_1,g_2,g_3[/imath] form a basis of [imath]R^3[/imath] we can represent [imath]v=ag_1+bg_2+cg_3[/imath] and:
[math](A-I)(ag_1+bg_2+cg_3) = 0 =[/math][math]= aAg_1+bAg_2+cAg_3 - a g_1 - bg_2 - cg_3[/math][math]= a\lambda_1 g_1 + b\lambda_2g_2+c\lambda_3g_3 - a g_1 - bg_2 - cg_3[/math]Since [imath]\lambda_1 = 1[/imath] the above can be rewritten as
[math]0 = b\lambda_2g_2+c\lambda_3g_3 - bg_2 - cg_3 = b(\lambda_2-1) g_2+c(\lambda_3-c)g_3[/math]But since [imath]g_2,g_3[/imath] are linearly independent and [imath]\lambda_{2,3} \neq 1[/imath] we have
[imath]b=c=0[/imath] and thus [imath]v=ag_1[/imath] contrary to our assumption.
Claim: If [imath]A^3=I[/imath] then either [imath]A-I=0[/imath] or the rank of [imath]A-I[/imath] is 2.
Proof: Since [imath]A^3 = I[/imath] then for all 3 eigenvalues we have [imath]\lambda_i^3=1[/imath]. Because [imath]A[/imath] is real this means that either [imath]\forall i: \lambda_i = 1[/imath] or we have [imath]\lambda_0=1[/imath] and [imath]\lambda_1,\lambda_2[/imath] are two other (complex) cube roots of 1.
If [imath]\forall i:\lambda_i = 1[/imath] then [imath]\forall v \in R^3 : Av = v \Longrightarrow A=I[/imath], in which case [imath]A-I=0[/imath] has rank 0.
If all [imath]\lambda_i[/imath] are different then we'll show that the rank of [imath]A-I[/imath] is 2. Let [imath]g_1, g_2, g_3[/imath] be eigenvectors of [imath]A[/imath], and [imath]Ag_1 = g_1[/imath] or [imath](A-I)g_1 = 0[/imath]. This means that the rank is less than 3. We prove that the rank is 2 by contradiction: assume that it is less than 2, which means that is another vector [imath]v[/imath] such that [imath]v\neq \alpha v[/imath] for which [imath](A-I)v=0[/imath], or [imath]Av=v[/imath].
Since [imath]g_1,g_2,g_3[/imath] form a basis of [imath]R^3[/imath] we can represent [imath]v=ag_1+bg_2+cg_3[/imath] and:
[math](A-I)(ag_1+bg_2+cg_3) = 0 =[/math][math]= aAg_1+bAg_2+cAg_3 - a g_1 - bg_2 - cg_3[/math][math]= a\lambda_1 g_1 + b\lambda_2g_2+c\lambda_3g_3 - a g_1 - bg_2 - cg_3[/math]Since [imath]\lambda_1 = 1[/imath] the above can be rewritten as
[math]0 = b\lambda_2g_2+c\lambda_3g_3 - bg_2 - cg_3 = b(\lambda_2-1) g_2+c(\lambda_3-c)g_3[/math]But since [imath]g_2,g_3[/imath] are linearly independent and [imath]\lambda_{2,3} \neq 1[/imath] we have
[imath]b=c=0[/imath] and thus [imath]v=ag_1[/imath] contrary to our assumption.