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Construct a linear operator $T$ with minimal polynomial $ x^2(x-1)^2 $ and characteristic polynomial $x^3(x-1)^4$. Describe the primary decomposition of the vector space under $T$ and find the projections on the primary components. Find a basis in which the matrix of $T$ is in Jordan form.Also find an explicit direct sum decomposition of the space into $T$-cyclic subspaces and give the invariant factors.

I have a problem constructing the operator. I tried to construct the matrix of an operator in $F^7$ which decomposes the space in $W_1$ (the null space of $T^2$) and $W_2$ (the null space of $(T-1)^2$). I tried to use the standard basis for $F^7$ and arbitrary dimensions for each subspace $W_1$ and $W_2$ but with $\dim W_1+\dim W_2=7$ but failed. As far as the projections I understand that $E_1=(T-1)^2g_1(T)$ and $E_2=T^2g_2(T)$ with $(x-1)^2g_1+x^2g_2=1$. I believe if the operator is constructed I can answer the other questions though it will be helpful if the whole answer is concentrated here!

Thank you in advance!

Marc van Leeuwen
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samer abdullah
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    Do you know that when restricting to the generalised eigenspace for $\lambda$, the characteristic and minimal polynomials of the restriction each retain only powers of $(x-\lambda)$ of the full characteristic respectively minimal polynomial? (Note $\lambda=0$ gives generalised eigenspace $W_1$, and $\lambda=1$ gives $W_2$.) The "characteristic" part of that statement will allow you to find $\dim W_1$ and $\dim W_2$. – Marc van Leeuwen Apr 25 '19 at 09:04
  • So W1 is 3 and W2 is 4 if I get this right?still though how do I construct the operator?Thanks for answering!!! – samer abdullah Apr 25 '19 at 13:02

1 Answers1

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Since you are allowed to choose $T$ and even your vector space yourself, you can simplify your task by working in $\Bbb C^7$ an choosing its matrix with respect to the standard basis it to be a Jordan normal form. The characteristic and minimal polynomials both tell you that $\lambda=0,1$ are the two eigenvalues, you know for $\lambda=0$ that the largest block is of size$~2$ and the sum of the sizes of blocks is $4$, and similarly you know for $\lambda=1$ that the largest block is of size$~2$ and the sum of the sizes of blocks is $3$. There is some choice for the remaining block sizes, at least for $\lambda=0$, but you may choose all of them to have size$~1$, which is simplest.

The primary decomposition of the vector space is that into generalised eigenspaces, each of which is spanned by successive vectors of the standard basis; the corresponding projections are obvious. The cyclic subspaces are those corresponding to the individual Jordan blocks.

Marc van Leeuwen
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