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Let $T:R^4\to R^4$ be a linear map. Null space of $T$ is {$(x,y,z,w)\in R^4:x+y+z+w=0$}. Rank of $(T-4I_4)$ is 3.The minimal polynomial of $T$ is $x(x-4)^a$. Then prove that $a=1$.

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    The minimal polynomial has to have degree 4, so your question should read 'The minimal polynomial is $x^{4-a}(1-x)^a$. Then proof that $a = 1$.' Or perhaps even: 'The minimal polynomial is $p(x)x(1-x)^a$ for some polynomial $p$ of degree $3 - a$. Then prove that $a = 1$'. – 2017-02-02
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    @ Vincent: the minimal polynomial is not the char. polynomial ! – 2017-02-02
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    O wait, now I see. Ok. – 2017-02-02
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    It is simply the case that the minimal polynomial must contain the factors at their lowest degree. If the minimal polynomial were $x(x-4)^2$ then the polynomial $x(x-4)$ would divide it, and this would be a contradiction. $a=0$ cannot be either otherwise $T$ would be equivalent to the identity matrix, and this cannot be since $(T-4I_4)$ has rank 3. – 2017-02-02

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Nullity of $A$ =$3$ as you can see from the definition.

Nullity of A is precisely the Geometric multiplicity of $0$. So that gives you G.M of $0= 3$

Given is rank of $(T-4I)=3$. This implies Nullity of $(T-4I)=1$ ,which is the Geometric multiplicity of $4$ is $1$.

So in total you get $4$ linearly independent eigenvectors i.e 3 eigenvectors corresponding to $0$ and $1$ eigenvector corresponding to $4$.

And if a linear operator has the number of linearly independent eigenvectors equal to the dimension of the domain(in this case $4=dim(R^4)$) then the operator is diagonalizable.hence it's minimal polynomial is product of linear or irreducible factors.Hence $a=1$

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    if an operator is diagonalizable. then it's minimal polynomial is product of linear or irreducible factors. Is it so? – 2017-02-02
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    yes you can add the word distinct linear factors if you want. – 2017-02-02
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Since the minimal polynomial vof $T$ is $x(x-4)^a$, we have that $T$ has exactly the eigenvalues $0$ and $4$ and the char. polynomial of $T$ is $x(x-4)^3$. Then we have

$\mathbb R^4=ker(T) \oplus ker((T-4I_4)^3)$

hence

$4= \dim ker(T)+ \dim ker((T-4I_4)^3)=3+\dim ker((T-4I_4)^3)$,

thus

$\dim ker((T-4I_4)^3)=1$, which shows that $ker((T-4I_4)^3)=ker(T-4I_4).$

Consequently: $\mathbb R^4=ker(T) \oplus ker(T-4I_4)$.

This gives $a=1$

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    I could not get one thing. How to say $ker(T-4I)^3=ker(T-4I)$ – 2017-02-02
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    How to say $ker(T-4I)^3=ker(T-4I)$ if their nullities are equal. – 2017-02-02
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    Another thing, char. poly. may be- $x^3(x-4)$ or $x^2(x-4)^2$ – 2017-02-02
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    We have $\{0\} \ne ker(T-4I) \subseteq ker(T-4I)^3$. Since $ \dim ker((T-4I_4)^3)=1$ and $\dim ker(T-4I_4) \ge 1$ we get $ker(T-4I)^3=ker(T-4I)$ – 2017-02-02