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Questions tagged [linear-algebra]

For questions on linear algebra, including vector spaces, linear transformations, systems of linear equations, spanning sets, bases, dimensions and vector subspaces.

Linear algebra is concerned with vector spaces and linear transformations between them:

$$(x_1, \dots, x_n)\to a_1x_n+\dots+a_nx_n$$

Concepts include systems of linear equations, bases, dimensions, subspaces, matrices, determinants, kernels, null spaces, column spaces, traces, eigenvalues and eigenvectors, diagonalization and Jordan normal forms.

This is a general tag, most of the subjects included have secondary tags, e.g.

127034 questions
11
votes
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If $A,B$ symmetric positive semidefinite, show tr$(AB) \geq 0$

Supposing $V$ is a finite dimensional vector space (over $\mathbb{R}$) of dimension $n$, and $A,B$ are symmetric positive definite linear mappings from $V$ to $V$, how can I show that in any orthonormal basis $\mathrm{tr}(AB) \geq 0$? I noticed that…
nullUser
  • 27,877
11
votes
1 answer

Dimension of the sum of subspaces

Let $V_1, \ldots, V_n$ be $n$ subspaces of a vector space $V$. Is there a formula for $\dim(V_1 + \cdots + V_n)$ similar to $\dim(V_1 + V_2)=\dim(V_1) + \dim(V_1) - \dim(V_1 \cap V_2)$?
Peter P.
  • 121
10
votes
1 answer

Matrix-version of $\frac1a+\frac1b=\frac{a+b}{ab}$?

In my Linear Algebra Textbook one exercise is to find the matrix analogue of $$ \frac1a+\frac1b=\frac{a+b}{ab} $$ my immediate response was $$ A^{-1}+B^{-1}=A^{-1}(A+B)B^{-1} $$ is that a reasonable answer or do someone have better…
String
  • 18,395
10
votes
2 answers

Are three matrices linearly independent and form a basis of $M_2(\mathbb R)$?

I know how to prove whether or not vectors are linearly independent, but can't apply the same thing to matrices it seems. Given three 2x2 matrices, for example: $$A = \begin {bmatrix} -1&1 \\\\ -1&1 \ \end{bmatrix}$$ $$B = \begin {bmatrix} 1&1 \\\\…
Ciarán Tobin
  • 203
10
votes
2 answers

Relating Linear Independence to Affine Independence

Question: (From an Introduction to Convex Polytopes) Let $(x_{1},...,x_{n})$ be an $n$-family of points from $\mathbb{R}^d$, where $x_{i} = (\alpha_{1i},...,\alpha_{di})$, and $\bar{x_{i}} =(1,\alpha_{1i},...,\alpha_{di})$, where $i=1,...,n$. Show…
Samuel Reid
  • 5,072
10
votes
1 answer

All linear combinations diagonalizable over $\mathbb{C}$ implies commuting.

Let $A, B \in \mathcal{M}_n(\mathbb{C})$ such that for all $x,y \in \mathbb{C}, xA+yB$ is diagonalizable. Show that $AB=BA$. My idea (not really an attempt) : It suffices to show that $A$ and $B$ are simultaneous diagonalizable. Then one can use…
user146010
10
votes
3 answers

Is there a similarity transformation rendering all diagonal elements of a matrix equal?

I'm especially interested in SL$(2,\mathbb C)$, i.e. $2\times2$ matrices with determinant one, in which case I'm looking for a transformation from $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ to $\begin{pmatrix}\frac{a+d}2&x\\ y&\frac{a+d}2\end{pmatrix}$…
Tobias Kienzler
  • 6,609
10
votes
1 answer

Question about the converse of a well known result from Linear Algebra

I am a graduate student studying for a Linear Algebra qualifying exam and I have been going over sample problems from previous exams. The recommended text for these problems are Hoffman and Kunze "Linear Algebra", Chapter three of Jacobson "Algebra…
user7980
  • 3,073
10
votes
2 answers

Prove that if $\operatorname{rank}(T) = \operatorname{rank}(T^2)$ then $R(T) \cap N(T) = \{0\}$

Let $V$ be a finite-dimensional vector space and let $T:V\to V$ be linear. Prove that if $\operatorname{rank}(T) = \operatorname{rank}(T^2)$, then $R(T) \cap N(T) = \{0\}$. I don't see this implication, at all. Please give hints and explain…
JohanLiebert
  • 1,127
10
votes
2 answers

Is the finite dimension required in this proof?

Let $V$ and $W$ be vector spaces over a field $K$. If a linear map $L:V \rightarrow W$ is surjective then its dual is injective. If $V$ and $W$ are finite dimensional then the converse holds, i.e. $L^*:W^* \rightarrow V^*$ injective implies $L$…
inquisitor
  • 1,740
10
votes
2 answers

Is there an example of an orthogonal matrix which is not diagonalizable

Based off the theory I can't see any reason that an example would not exist. Specifically, the fact that A matrix is orthogonal only implies that the possible eigenvalues are $\pm 1$. However, we don't know anything about the sizes of the…
jacob cohen
  • 101
10
votes
2 answers

Prove that if $A$ is diagonalizable then there is a matrix $B$ such that $B^{2012} = A$

Given: $$A \in M_{n\times n} (\mathbb C) \; , \; A \; \text{is diagonalizable}$$ We need to prove that: $$ \exists B \in M_{n\times n} (\mathbb C) \; : B^{2012} = A$$ What I said so far: If $A$ is diagonalizable, then $\exists P$ and $D$ such that…
TheNotMe
  • 4,841
10
votes
1 answer

What is the difference between finding a basis for a complex and a real space?

Given that $V = \mathbb{C}^{3}$ how does one determine if a set of vectors $A \subset V$ is a basis? Am I allowed to assume that the the field is now $\mathbb{C}$ and so I am allowed to use all $\lambda \in \mathbb{C}$ as scalars? If so, $E = \{…
ghshtalt
  • 2,753
10
votes
1 answer

Do complex eigenvalues for planar systems with a real valued matrix always have complex-valued eigenvectors?

sorry if this is a dumb question, but I was doing some homework and I noticed that whenever I solved a planar system which had complex eigenvalues, I would always end up with complex eigenvectors. I was wondering whether I could ever somehow get a…
Brodrick
  • 103
10
votes
3 answers

Burning a rope to count time

A rope burns irregularly in 16 minutes and costs 32 rupees, while a second rope burns also irregularly in 7 minutes and costs 14 rupees. Both can be lit only at one end and can be turned off and lit again as many times we want, until they are…
Abhishek Bakshi
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