It seems that a Jordan basis, for, say, an operator $φ:V→V$ with $V$ a vector space can have several Jordan bases. However I don't see how this is true and I didn't achieve to prove it. Indeed I tried assuming the uniquness of a Jordan basis of $\varphi$ to achieve a contradiction but I did not find any.. I also looked up on mathstack but the only thing I found was this:  And it doesn't realy help because I'm more looking for the general case, which is unclear to me. Could anyone explain why this is true and help me proving it ?
123998
Is Jordan basis unique? I have a 4x4 matrix, find eigenvectors and one generalized eigenvector, also trying different linearly independent eigenvectors but the matrix P so that PJP^-1 =A only works for certain vector. \begin{pmatrix}-\frac{1}{2}&amp;0&amp;-\frac{1}{2}&amp;0\\ \frac{3}{2}&amp;0&amp;\frac{1}{2}&amp;0\\ -\frac{3}{2}&amp;0&amp;-\frac{3}{2}&amp;-1\\ 1&amp;1&amp;1&amp;1\end{pmatrix}\begin{pmatrix}1&amp;-1&amp;0&amp;-1\\ 0&amp;2&amp;0&amp;1\\ -2&amp;1&amp;-1&amp;1\\ 2&amp;-1&amp;2&amp;0\end{pmatrix}\begin{pmatrix}1&amp;1&amp;0&amp;0\\ -1&amp;0&amp;1&amp;1\\ -3&amp;-1&amp;0&amp;0\\ 3&amp;0&amp;-1&amp;0\end{pmatrix}=\begin{pmatrix}-1&amp;0&amp;0&amp;0\\ 0&amp;1&amp;0&amp;-1\\ 0&amp;0&amp;1&amp;1\\ 0&amp;0&amp;0&amp;1\end{pmatrix}
114380
Let f,g be bilinear forms on a finite dimensional vector space.  (a) Suppose g is non-degenerate. Show tha there exist unique linear operators T1, T2 on V such that f(a,b)=g(T1a,b)=g(a,T2b) for all a,b.  (b) Show that this result may not be true if g is degenerate.  My attempt:  Although I am clueless about this question, I was wondering if it has something to do with the adjoint of a linear operator. If we have a linear operator T on a hermitian space V then = and = for all v,w in V. But then how would you relate it to the other bilinear form and how would you use the fact that one of the bilinear forms is non-degenerate?
74001
If $X$ is a vector space over $\mathbb R$ and $B=\{x_i; i\in I\}$ is a Hamel basis for $X$, then for each $i\in I$ we have a linear functional $a_i(x)$ which assigns to $x$ the $i$-th  coordinate, i.e., the functions $a_i$ are uniquely determined by the conditions that $$x=\sum_{i\in I} a_i(x)x_i,$$ where only finitely many summands are non-zero.  If $X$ is a Banach space, then at most finitely many of them can be continuous.  I have learned the following argument from comments in .     Suppose that $\{b_i; i\in\mathbb N\}$ is an infinite subset of $B$ such that each $f_{b_i}$ is continuous. W.l.o.g. we may assume that $\lVert{b_i}\rVert=1$.      Let   $$y:=\sum_{i=1}^\infty \frac1{2^i}b_i.$$   (Since $X$ is complete, the above sum converges.)      We also denote $y_n:=\sum_{i=1}^n \frac1{2^i}b_i$. Since $y_n$ converges to $y$, we have $f_{b_k}(y)=\lim\limits_{n\to\infty} f_{b_k}(y_n)=\frac1{2^k}$ for each $k\in\mathbb N$. Thus the point $x$ has infinitely many non-zero coordinates, which contradicts the definition of Hamel basis.   I have stumbled upon  in the book Christopher Heil: A Basis Theory Primer. Springer, New York, 2011. In this exercise we are working in an infinite-dimensional space $X$. Basically the same notation as I mentioned above is introduced, $a_i$'s are called coefficient functionals and then it goes as follows:     (a) Show by example that it is possible for some particular functional $a_i$ to be continuous.      (b) Show that $a_i(x_j) = \delta_{ij}$ for $i,j\in I$.      (c) Let $J = \{i\in I : a_i \text{ is continuous}\}$. Show that $\sup \{j\in J; \lVert a_j \rVert&lt;+\infty\}$      (d) Show that at most finitely many functionals $a_i$ can be continuous, i.e., $J$ is finite.      (e) Give an example of an infinite-dimensional normed linear space that has a Hamel basis $\{x_i; i\in I\}$ such that each of the associated coefficient functionals $a_i$ for $i\in I$ is continuous.   The part (c) can be shown easily using  (a.k.a. Uniform boundedness principle). But I guess that the author of the book has in mind a different proof for part (d) from what I sketched above, since (c) is an easy consequence of (d) -- so he would probably not be put the exercises in this order. (But maybe I was just trying to read to much between the lines.)  Question: I was not able to find a proof od (d) which uses (c). Do you have some idea how to do this?  NOTE: My question is not about the parts (a), (b), (e). I've included them just for the sake of the completeness, in order to include sufficient context for the question.
115553
I would appreciate a hint on the following problem:     Let $V$ be a finite dimensional vector space over $F$. There are two scalar products such that: $$ \forall \ w,v \in V \ \Big(\langle v,w\rangle_1=0 \implies \langle v,w\rangle_2=0\Big) $$ Show that $$ \exists \ c \in F \ \ \forall \ w,v \in V \ \Big(\langle v,w\rangle_1=c \langle v,w\rangle_2\Big) $$   I have tried to define an orthonormal basis with respect to $\langle \cdot,\cdot\rangle_1$ hoping that the transformation matrices would be different by a constant, yet it seems to lead nowhere.
119326
In my preparations for an upcoming exam, I'm working through the past exams available from my department.  One of the questions asks for an example of a linear operator on some vector space with more than one right inverse.  In another part of the same problem, I proved that right-inverse uniqueness implies invertibility, but I feel like I'm missing how that result will help me find or construct an example (assuming that it's applicable.)  I've considered various square matrices (because I'm looking for an operator) over various fields, but nothing has worked yet and I don't feel like I'm going about it the right way.
24000
So far, I understand that I have to look at each Jordan block. How do I prove that for a Jordan block with a value of x, $J_n(x)$, the Jordan decomposition of $(J_n(x))^2$ is $J_n(x^2)$ ? I have a feeling this will lead me to the solution. Thanks for your time.
22385
I think the answer is yes.   Sketch of the proof Consider $\mathbb{R}$ as a vector space over $\mathbb{Q}$. Let $\{e_\lambda:\lambda\in\Lambda\}\subset\mathbb{R}$ be its Hamel basis. Then $\{(e_{\lambda_1},e_{\lambda_2}):\lambda_1,\lambda_2\in\Lambda\}\subset\mathbb{R}^2$ is a Hamel basis of $\mathbb{R}^2$. Consider set theoretic bijection $i:\Lambda\to\Lambda\times\Lambda$ define map $\varphi:\mathbb{R}\to\mathbb{R}^2$ by its action on element of Hamel basis of $\mathbb{R}$ by equality $\varphi(e_\lambda)=(e_{i(\lambda)_1},e_{i(\lambda)_2})$. This is an isomorphism between $\mathbb{Q}$-vector spaces $\mathbb{R}$ and $\mathbb{R}^2$.  Question Could you tell me is this proof correct, and if it is not, where is the mistake?  EDIT: Thanks to GAJO we found that this proof is wrong. The correct one basis is  $$ \{(e_\lambda,0):\lambda\in \Lambda\}\cup\{(0,e_\lambda):\lambda\in\Lambda\} $$ So we had to look at bijections of the form $i:\Lambda\to\Lambda\coprod\Lambda$ and the desired linear operator is defined by equalities $\varphi(e_\lambda)=(e_{i(\lambda)},0)$ if $i(\lambda)$ lies in the first copy $\Lambda$ and $\varphi(e_\lambda)=(0,e_{i(\lambda)})$ otherwise.
5625
The Jordan and Frobenius normal forms of a linear map $A:\Bbb R^n \rightarrow \Bbb R^n$ seem to be maximally simple representations of $A$ in the sense that one of them contains as few nonzero entries as possible. But how do you prove that?   More precise, show that for every $A:\Bbb R^n \rightarrow \Bbb R^n$ and every Basis $B$ of $\Bbb R^n$, the transformation matrix $_B A _B$ has at least as many nonzero entries as the Jordan normal form or the Frobenius normal form of $A$ or, otherwise, give a counterexample.
108113
For a linear transformation $T$ on a vector space $V$ of dimension $n .$ Suppose it is given that for some vector $\mathbf{v},$ the vectors $\mathbf{v}, T(\mathbf{v}), T^{2}(\mathbf{v}), \ldots, T^{n-1}(\mathbf{v})$ are linearly independent, then is it true that every eigenvalue of $T$ has only one corresponding eigenvector up to a scalar multiplication.
25928
Suppose I have an eigenvalue $\lambda$, now I want to determine the number of Jordan blocks corresponding to that eigenvalue, as well as size of each block. I know that:  number of blocks is equal to the number of eigenvectors there are $\dim \ker (A - \lambda I)^k - \dim \ker (A - \lambda I)^{k-1}$ blocks of size at least $k$ Why exactly are those statements true. Could someone provide me with a proof or at least some reasoning behind this?
109497
Suppose we have $n$ linearly independent vectors $\mathbf{v}_{1}\ldots\mathbf{v}_{n}$ in $\mathbb{R}^{n}$. I know that they do span $\mathbb{R}^{n}$, because we can easily specify a non-singular map which sends the $\mathbf{v}_{i}$s to the standard basis, and then to whichever vector in $\mathbb{R}^{n}$ we choose.    My question is: do we need all the machinery of linear maps, determinants, etc. or is there a proof which is closer to the definitions? Every time I start writing down a proof I end up wanting to say "and this set of equations can be solved uniquely because this matrix is non-singular". Is this necessary?
1217
Seems like I , I think I am missing something important.     Let $V$ be an $n$ dimensional inner product space ($n \geq 1$), and $T\colon\mathbf{V}\to\mathbf{V}$ be a linear transformation such that:         $T^2 = T$   $||T(a)|| \leq ||a||$ for every vector $a$ in $\mathbf{V}$;         Prove that a subspace $U \subseteq V$ exist, such that $T$ is the orthogonal projection on $U$.   Now, I know these things:   The fact hat $T^2 = T$ guarantees that $T$ is indeed a projection, so I need to prove that T is an orthogonal projection (I guess this is where $||T(a)|| \leq ||a||$ kicks in). To do this I can prove that:   For every $v$ in $ImT^{\perp}$, $T(v) = 0$ Alternatively, I can prove that for every $v$ in $ImT$ and $u$ in $KerT$, $(v,u)=0$. $T$ is  (according to Wikipedia) The matrix $A = [T]_{E}$ when $E$ is an orthonormal basis, is hermitian (this is equivalent to the previous point). What else?    I've been thinking about it for quite some time now, and I'm pretty sure there is something big I'm missing, again. I just don't know how to use the data to prove any of these things.  Thanks!
44655
Reproduced from Exercise 2.2 of Nielsen &amp; Chuang's Quantum Computation and Quantum Information (10th Anniversary Edition):     Suppose $V$ is a vector space with basis vectors $|0\rangle$ and $|1\rangle$, and $A$ is a linear operator from $V$ to $V$ such that $A|0\rangle = |1\rangle$ and $A|1\rangle = |0\rangle$. Give a matrix representation for $A$, with respect to the input basis $|0\rangle, |1\rangle$, and the output basis $|0\rangle, |1\rangle$. Find input and output bases which give rise to a different matrix representation of $A$.   Note: This question is part of a series attempting to provide worked solutions to the exercises provided in the above book.
102940
I need to show that $v_1,...,v_n$ is basis for $V$ whenever $e_1,...,e_n$ is an orthonormal basis for V and $v_1,...,v_n$ are vectors in $V$ such that $$\left\Vert e_i-v_i\right\Vert &lt; \frac{1}{\sqrt{n}}.$$ I know already that I only need to prove that they are lineary independent. Also I tried to prove it by contradiction, but I got stuck fairly fast. Any help, or hint how to prove it (with ot without contradiction) will be appriciated.
33091
If $V_0$ is the subspace consisting of matrices of the form $C=AB-BA$ for some $A,B$ in a vector space $V$ then $V_0=\{A\in V|\operatorname{Trace}(A)=0\}$.  The problem above is one of the past qualifying exam problems. I can prove that  $V_0\subset\{A\in V|\operatorname{Trace}(A)=0\}$ but I do not know what to do with converse.  Thank you in advance.
121835
I believe I am very close to finishing this proof, but I cannot figure out the last part. If anybody could check my work and maybe give me a little hint, it would be greatly appreciated!  Let $V$ be a finite-dimensional vector space and $T \in \mathcal{L}(V)$, and let $\mathbf{u,v} \in V$ be eigenvectors of $T$.  Claim:  If $\mathbf{u} + \mathbf{v}$ is an eigenvector of $T$, then $\mathbf{u}, \mathbf{v}$, and $\mathbf{u+v}$ all correspond to the same eigenvalue.  Proof (So far!): Suppose $T(\mathbf{u}) = \lambda_1 \mathbf{u}$ and $T(\mathbf{v}) = \lambda_2 \mathbf{v}$ with $\mathbf{u}, \mathbf{v} \neq \mathbf{0}$.   Now suppose $T(\mathbf{u+v}) = \lambda_3(\mathbf{u+v})$ with $\mathbf{u+v}\neq \mathbf{0}$.  Then, $$ T(\mathbf{u}) + T(\mathbf{v}) = \lambda_3 \mathbf{u} + \lambda_3 \mathbf{v}\\ \lambda_1\mathbf{u} + \lambda_2\mathbf{v} =  \lambda_3 \mathbf{u} + \lambda_3\mathbf{v}\\ \lambda_1\mathbf{u} + \lambda_2\mathbf{v} - \lambda_3 \mathbf{u} -\lambda_3\mathbf{v} = \mathbf{0}\\ (\lambda_1 - \lambda_3) \mathbf{u} + (\lambda_2 - \lambda_3)\mathbf{v} = \mathbf{0} $$  Now I know in order to show that $\lambda_1 = \lambda_2 = \lambda_3$, I must show that the only solution to the last line is the trivial one. This would imply that $\mathbf{u}$ and $\mathbf{v}$ are linearly independent which I am unconvinced of!  The only information I have to my advantage I haven't used yet is the fact that $\mathbf{u}, \mathbf{v}, \mathbf{u+v} \neq \mathbf{0}$. I really cannot see how this information can help me though.  Perhaps I am going about this wrong, but that's why I want to ask! Thanks for your help.
63028
Suppose we have vector spaces $\mathbb W$ and $\mathbb V$.  A given basis for $\mathbb W$ is $\{(1, 1, 0, -1), (0, 1, 3, 1)\}$, and a given basis for $\mathbb V$ is $\{(1, 2, 2, -2), (0, 1, 2, -1)\}$.  How to find a basis for $\mathbb W\cap\mathbb V$?