This is to show that the minimal polynomial does not determine the jordan normal form. I've tried a few but haven't got anywhere.  Why is this marked as duplicate? The other post does not give me an answer.
75934
Why and how is the  of a matrix in $M_3(\mathbb C)$ fully determined by its characteristic and minimal polynomials? And why does it fail for $n &gt;3$?   Thanks.
84412
I'm having a lot of trouble understanding the process of finding a basis for the Jordan canonical form (the "algorithm"). My textbook (Friedberg 4E) isn't very clear, and I can't seem to find anything online.   If we consider the example $A$ = $\begin{bmatrix}6 &amp; -2 &amp; -1\\3 &amp; 1 &amp; -1\\2 &amp; -1 &amp; 2\end{bmatrix}$, we get the characteristic polynomial $p(t)$ = ($3 - t$)$^3$.  I understand the process of finding the matrix $J$ in $A = P^{-1}JP$, but I just can't seem to grasp finding $P$. Any clarification/explanation on the steps to do so would be incredibly helpful. Thank you.
55879
The following is an exercise (Exercise #3 (a), Chapter 3, page 28) from Richard Stanley's Algebraic Combinatorics.  Let $P(x)$ be a nonzero polynomial with real coefficients.  Show that the following two conditions are equivalent:  There exists a nonzero polynomial $Q(x)$ with real coefficients such that all coefficients of $P(x) Q(x)$ are nonnegative.  There does not exist a real number $a &gt; 0$ such that $P(a) = 0$.    That the first item implies the second is straightforward (since if $a$ is a positive real root of $P(x)$ then it is a positive real root of $P(x)Q(x)$, but since $P(x)Q(x)$ is nonzero with all coefficients nonnegative, this is impossible).  However, I can't seem to find a way to prove that the second item implies the first.  I would very much like to see a proof if someone can provide it.
116860
Calculate the minimal polynomial of $\alpha=\sqrt 2e^\frac{2\pi i}{3}$ over $\mathbb Q$.  My try:  Let $x=\sqrt 2e^\frac{2\pi i}{3}\implies x-\sqrt 2(-\frac{1}{2}+i\frac{\sqrt 3}{2})=0$.   On solving completely I got the polynomial to be $x^4+2x^2+4$ which is satisfied by $\alpha$ which is a monic polynomial .Also it is an irreducible polynomial over $\mathbb Q$.  However I am finding it difficult to check that is the minimal polynomial i.e it is the lowest degree polynomial satisfied by $\alpha $ over $\mathbb Q$.  How should I do it?
77261
I'm working on the following problem:     Let $A \in M_3 (\mathbb {R})$ be such that $A^8=I$. Then           the minimal polynomial of $A$ can only be of degree $2$.     the minimal polynomial of $A$ can only be of degree $3$.     either $A = I$ or $ A = -I$.     there are uncountably many such $A$.        By taking $A=I $ we can eliminate options (1) &amp; (2). For the minimal polynomial of $A$ in that case is of degree $1$. Now take   $$A = \begin{bmatrix}      1 &amp; 0 &amp; 0 \\      0 &amp; -1 &amp; 0 \\      0 &amp; 0 &amp; -1 \end{bmatrix}$$  Then $A^8=I $ but $A$ is neither $I$ nor $-I$. So option (3) is eliminated. Now I don't know how to proceed about with option  (4). Can someone help me please? Thanks in advance.
63576
Yet again I find myself stuck on a Spivak question. This time it is simply the question that isn't clear to me. It states:  If $x_1, ..., x_n$ are distinct numbers, find a polynomial function $f_i$, of degree $n-1$ which is 1 at $x_i$ and 0 at $x_j$ for $j\neq i$. Hint: the product of all $(x-x_j)$ for $j\neq i$, is 0 at $x_j$ if $j\neq i$. This product is usually denoted by: $$\prod_{\begin{smallmatrix}{j=1}\\ {j\neq i} \end{smallmatrix}}^n(x-x_j)$$  To be honest I don't know where to begin. I understand how to create a polynomial equation for a given set of results but this is quite strange. I'm not sure exactly what the author is expecting.  The answer book shows:  $$f_i(x)={\prod_{\begin{smallmatrix}{j=1}\\ {j\neq i} \end{smallmatrix}}^n(x-x_j)}/{\prod_{\begin{smallmatrix}{j=1}\\ {j\neq i} \end{smallmatrix}}^n(x_i-x_j)}$$  Which I have expanded out just fine but don't know what to do with.
20168
I am currently reading two linear algebra books. One is Hoffman/Kunze's and the other one is Friedberg/Insel/Spence's.  They define Jordan canonical form of linear operator in different ways.  In Hoffman's book, Jordan forms are obtained by the primary decomposition and cyclic decomposition. In this case, 1's can appear some sub-diagonal entries. For example,  $M = \begin{array}{cc} 2 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\ 1 &amp; 2 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 3 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1 &amp; 3 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; 3 \\ \end{array}$  But in Friedberg's (And almost any other linear algebra books for engineers..), 1's can appear some super-diagonal entries. For example,  $M' = \begin{array}{cc} 2 &amp; 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 2 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 3 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 3 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; 3 \\ \end{array}$  Can someone explain why they define differently and relation of each other?
94994
Let $M$ be a (real) $n \times n$ matrix. For $1 \leq i, j \leq n$ we denote by $M_{ij}$ the $(n-1) \times (n-1)$ matrix that we get when the $i$th row and $j$th column of $M$ are removed. Now, consider fixed $i$ and $j$ with $i\neq j$. Let $N$ be the $(n-2) \times (n-2)$ matrix that we get when removing both the $i$th and $j$th row and the $i$th and $j$th column from $M$. Then the following identity holds: $$ \det M \det N = \det M_{ii}\det M_{jj} - \det M_{ij} \det M_{ji}. $$  We were able to prove this looking at all different terms that can occur on both sides when evaluating the determinant as a sum of $n!$ terms (for instance, terms containing $a_{ji}a_{ij}$ are counted both $1$ time in the LHS and $1$ time in the RHS).  We are looking for a slick proof that does not involve writing out the determinant.  Any suggestions or approaches to this problem are appreciated!
70708
I would like to verify the validity of the following line of thought:  Let $K \subset E$ be a field extension. Let $K[x]$ be the polynomial ring over $x$ and denote $K(x)$ its field of quotients (the rational functions over $K$).  Now assume $f(x)=g(x)h(x)$ with $f(x),g(x) \in K[x]$ and $h(x) \in E[x]$ and all $f$, $g$, $h$ are different than zero. Then $\frac{f(x)}{g(x)} = h(x)$ in $E(x)$. So $h(x) \in K(x)$. But $h(x)$ is a polynomial, so $h(x) \in K[x]$.  Is there any other simpler argument than that?  Added:  Let me be more specific. Let $x^p - a \in K[x]$, with $p$ being prime and $char(K)=p$. Let $b$ be the unique root of $x^p - a$ in its splitting field $K(b)$. Then $x^p - a=(x-b)^p \in K(b)[x].$ Let $m_b(x) \in K[x]$ be the minimal polynomial of $b$ over $K$. Then $m_b(x)$ divides $(x-b)^p$ and so $m_b(x)=(x-b)^l$ and $l$ is the smallest positive integer for which $(x-b)^l \in K[x]$. Let $p = lq+r$ where $0 \ge &lt;l$. Then $x^p-a=(m_b(x))^q (x-b)^r$. Can i conclude that $(x-b)^r \in K[x]$ and so $r=0$ by the minimality of $l$?  Thanks.
76337
Like the title says, can a matrix in $\mathbb{R}$ have a minimal polynomial that has coefficients in $\mathbb{C}$?  I have a feeling the answer is yes because the characteristic polynomial can have coefficients in $\mathbb{C}$ and so I don't see any reason why not, I was merely asking in case there was some sneaky proof that proves that this cannot happen.
90035
remained somehow incompletely solved. The OP also asked for an explicit form of the minimal polynomial of the sum (respectively, product) of two algebraic elements (in a field extension). Although such form is, in general, impossible to find, he proposed the polynomial  $\Pi_j\Pi_i (x-\alpha_i-\beta_j)$ (respectively, $\Pi_j\Pi_i (x-\alpha_i\beta_j)$) as possible candidates.   The question is the following:      Have these two polynomials the coefficients in the base field?   I find it also useful for this .
77729
From Hoffman and Kunze.     Let $N_1$ and $N_2$ be 6 X 6 nilpotent matrices over the field F. Suppose that   $N_1$ and $N_2$ have the same minimal polynomial and the same nullity. Prove that   $N_1$ and $N_2$ are similar. Show that this is not true for 7 X 7 nilpotent matrices.   Right so nilpotent matrices have $N^k= 0$ for some k. Since these have the same nullity and minimal polynomial $N_1^k = N_2^k = 0$ from some k right? The same minimal polynomial implies the same characteristic values but thats not enough to say they have the same jordan form and are thus similar right?  and how would I get started in proving this breaks for 7x7 nilpotent matrices?  Thanks in advance for any assistance.
39724
Suppose that $\hat{x}$ is the least squares solution of $Ax = b$.  That is, $\min{|r|}, r = Ax - b$ with respect to $x$.  Is $\hat{x}$ also the least squares solution of $JAx = Jb$?  I suspect so, but I'm not sure how to prove it.  If $J$ is invertible, it's easy enough to prove as you can just cancel it out from both sides.  But what if it's not invertible?
21031
While reading this  I do not understand that how can we claim that there exist a polynomial with the properties, even if it exist then how is it same for the function and the inverse.
97733
I am attempting to compute the minimal polynomial of $1 + 2^{1/3} + 4^{1/3}$ over $\mathbb Q$. So far, my reasoning is as follows:   The Galois conjugates of $2^{1/3}$ are $2^{1/3} e^{2\pi i/3}$ and $2^{1/3} e^{4\pi i /3}$. We have $4^{1/3} = 2^{2/3}$, so the image of $4^{1/3}$ under an automorphism $\sigma$ fixing $\mathbb Q$ is determined by the image of $2^{1/3}$: it must equal the square of $\sigma(2^{1/3})$. Therefore, the Galois conjugates of $1 + 2^{1/3} + 4^{1/3}$ are $1 + 2^{1/3} e^{2\pi i/3} + 4^{1/3} e^{4\pi i/3}$ and $1 + 2^{1/3} e^{4\pi i/3} + 4^{1/3} e^{2\pi i/3}$. Therefore, the minimal polynomial is $(x-a)(x-b)(x-c)$, where  $$\begin{align*} a&amp;=1 + 2^{1/3} + 4^{1/3},\\ b&amp;=1 + 2^{1/3} e^{2\pi i/3} + 4^{1/3} e^{4\pi i/3},\text{ and}\\ c&amp;=1 + 2^{1/3} e^{4\pi i/3} + 4^{1/3} e^{2\pi i/3}. \end{align*}$$  However, this polynomial does not seem to have coefficients in $\mathbb Q$! What am I doing wrong?
49801
Can someone please give me a tip how to show that for distinct prime $p,q$ that $\mathbb{Q}(\sqrt{p},\sqrt[3]{q})=\mathbb{Q}(\sqrt{p}\sqrt[3]{q})$ and how to find the minimal polynomial of $\sqrt{p}\sqrt[3]{q}$ over $\mathbb{Q}$ ?
54881
I am told that $K \leq L$ is a field extension of degree 2 and that $K$ has characteristic not 2. I am then asked to show that $\exists \: \alpha \in L$ such that $L = K(\alpha), \; \alpha^2\in K $  So far I have that:  $\exists \; \alpha_1, \alpha_2 \in L$ such that $L = K( \alpha_1, \alpha_2)$  $\Rightarrow |L:K| = |L:K(\alpha_1)| |K(\alpha):K| $ so $|K(\alpha):K| = 1$ or $2$  $|K(\alpha_1):K| = 2 \Rightarrow L = K(\alpha_1)$ and $deg(f_{\alpha_1}(t)) = 2  $ where $f_{\alpha_1}(t)$ is the minimal polynomial of $\alpha_1$ over $K$   So $f_{\alpha_1}(t) = t^2 + at + b$ for some $a,b \in K$  Also, $|K(\alpha_1):K|=1 \Rightarrow K(\alpha_1) = K$ and $L = K(\alpha_2)$ which means $degf_{\alpha_2}(t) = 2$ where $f_{\alpha_2}(t)$ is the minimal polynomial of $\alpha_2$ over $K$.   So $f_{\alpha_2}(t) = t^2 + at + b$ for some $a,b \in K$  Given that we consider the same polynomial each time, relabel to consider:   $f_{\alpha}(t) = t^2 + at+ b $ for some $a,b \in K$  We have already demonstrated that $L=K(\alpha)$ for some $\alpha \in L$ so now we much show that $\alpha^2 \in K$  $\alpha^2 \in K \Leftrightarrow a = 0_K$  I now have no idea how to show that this coefficient of $f_{\alpha}(t)$ is necessarily $0$.   I know that $f_{\alpha}(t)$ has two roots in $L$ and that because $K$ and thus $L$ has characteristic not 2, they cannot be the same root, but I don't know how to then conclude that they must sum to $0$  Any help would be greatly appreciated.