Can anyone check these true and false statements about linear algebra?

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  1. For any square matrix $A$, the image of $A^7$ is contained in the image of $A$

    • I think this question is asking If $A^7x=b$, then $b$ must be in $A$ with some vector $y$ such that $Ay=b$. It Seems like "False". Can't come up with any good reasoning though.
  2. Every inner product space has an orthonormal basis.

    • By Gram-Schmidt process, it is true for finite dimensional space, but is it true for infinite dimensional?
  3. If $U$ and $W$ are subspaces of a finite dimensional vector space $V$ and $V = U + W$, then $\dim V ≤ \dim U + \dim W$.

    • we know that $\dim V=\dim(U+W)$ and $\dim(U+W) >\ dimU + \dim W$, and therefore $\dim V > \dim U +\dim U$. I think this is false.
  4. If $A$ is a $3 \times 3$ matrix which gives a rotation about some line through the origin in $\Bbb R^3$, then the columns of $A$ form a basis for $\Bbb R^3$.

    • Really have trouble with this question. Can't think of any valid argument or proof for this.
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  1. The image of $A$ consists of all columns of the form $Ab$, for whatever $b$ you want. Since $A^2b=A(Ab)$, you get that any vector in the image of $A^2$ is also in the image of $A$.

  2. No, it's not generally true for infinite dimensional spaces. For example, an infinite dimensional Hilbert space has a maximal orthogonal system such that the span is dense, so no vector can be orthogonal to all vectors in the system. However, the maximal orthogonal system is not a basis.

  3. $\dim(U+W)=\dim U+\dim W-\dim(U\cap W)$

  4. A rotation is bijective, so its matrix has rank $3$; hence…

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For #4, don't get intimidated by the statement. It is simply telling you that you have an orthogonal operator (or matrix w.r.t. to some basis), and we know that the columns (or rows) of any orthogonal (unitary, if the ground field is complex) matrix form an orthonormal basis for the vector space, which in this case is $R^3$.

(You may want to review which operators are considered orthogonal / unitary.)