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I was told that i can make a matrix A with vectors that span the subspace W which makes my the column space of my matrix equal to W. Then I can take the Null Space of A transpose which gives me (1,2,3,1). Does this mean the dim is one and it's automatically a basis because there is one vector?

  • $\begingroup$ How can a matrix be equal to a vector space? $\endgroup$ – José Carlos Santos Mar 30 '19 at 20:22
  • $\begingroup$ i meant to say the column space of the matrix would equal W $\endgroup$ – Samurai Bale Mar 30 '19 at 20:45

You have line spanned by the vector $v=[2,1,-6,18]^t.$ Then $$W^{\perp } = \{w \in \mathbb{R}^4: w\cdot v=0 \}\\ =\{w\in \mathbb{R}^4:2w_1+w_2-6w_3+18w_4=0\}.$$ If you fix $w_2 = \alpha, w_3= \beta, w_4 = \gamma$ then $$\begin{bmatrix} w_1 \\ w_2\\ w_3\\ w_4 \end{bmatrix} =\begin{bmatrix} -\alpha/2+3\beta-9\gamma \\ \alpha\\ \beta\\ \gamma \end{bmatrix}= \alpha\begin{bmatrix} -1/2 \\ 1\\ 0\\ 0 \end{bmatrix} + \beta \begin{bmatrix} 3 \\ 0\\ 1\\ 0 \end{bmatrix} + \gamma \begin{bmatrix} -9 \\ 0\\ 0\\ 1 \end{bmatrix}.$$ So the space $W^{\perp}$ is spanned by the vectors $$(-1/2,1,0,0), (3,0,1,0), (-9,0,0,1)$$ and its dimension is $3.$

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  • $\begingroup$ How would I construct a basis for W? $\endgroup$ – Samurai Bale Mar 30 '19 at 21:27
  • $\begingroup$ The basis of $W$ is simply the vector spanning it $(2,1,-6,18).$ $\endgroup$ – nls Mar 30 '19 at 21:27
  • $\begingroup$ I was given this explanation, If you have a basis for W, and take A to be the matrix with columns span those basis vectors, then the column space of A is W. Thus the perpendicular complement of W is exactly the perpendicular complement of Col(A), which we know is Null(A) Transpose. Wouldn't this yield a different answer? $\endgroup$ – Samurai Bale Mar 30 '19 at 21:29
  • $\begingroup$ @SamuraiBale It in fact yields the same answer, up to sign. $\endgroup$ – amd Mar 30 '19 at 21:33
  • $\begingroup$ how would this matrix look like? $\endgroup$ – Samurai Bale Mar 30 '19 at 21:45

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