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  1. If A is a 3 by 3 matrix which gives a rotation about some line through the origin in R^3 , then columns of A form a basis of R^3

  2. For any matrix A, the image of A^7 is contained in the image of A

  3. Every inner product space has an orthonormal basis.

Above are true or false statements.

Can anyone help me with whether these are true or false?

I have no idea about 1 and 2

For 3, every finite inner product space has an orthonormal basis by gram-schimith process,

does it hold for infinite space?

user197971
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  • if $A$ is a 'rotation' matrix it can alternatively be seen as a change of basis in which the vectors $e_1=(1,0,0),e_2=(0,1,0),e_3=(0,0,1)$ are mapped to the columns $e_1',e_2',e_3'$ of $A$ – obataku Dec 13 '14 at 06:05
  • This does not constitute a proof, but a matrix $A$ takes a vector $x$ and transforms it into a new vector $y$ by means of matrix multiplication, that is $Ax = y$. In that sense, $A^{7}$ takes whatever operation $A$ did, and applies it seven times. e.g., if a matrix $2\times 2$ matrix A reflects a vector about the $x$ axis, then $A^{7}$ reflects the vector 7 times which has a net result of reflecting it once. – JessicaK Dec 13 '14 at 06:16

1 Answers1

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Some Hints:

  1. If the columns of the $3 \times 3$ matrix $A$ do not form a basis of $\mathbb{R}^3$, then the columns are linearly dependent, and hence, $A$ has a non-trivial nullspace, i.e. there is a non-zero vector $x$ such that $Ax = 0$. Is this possible if $A$ is a rotation matrix?

  2. For any vector $x$ in the image of $A^7$, we have $x = A^7y$ for some vector $y$. To show that $x$ is also in the image of $A$, we need to write $x = Az$ for some vector $z$. Can you see how to do this?

  3. See this question.

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JimmyK4542
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