I am solving this problem 19 ( Sorry for uploading picture i ain't that good with latex. please help me out ! )
Here rank of Matrix is two so answer will be (b) or (d) .
I am trying to visualise this question using space xyz and $\hat{i}$,$\hat{j}$,$\hat{k}$ unit vectors along them respectively.
Now if I consider M as a matrix for linear transformation relative to some basis say B then V is just range space of M . So V is a plane in space but which plane ? If I keep changing B then I should be able to get every plane ( or not ?). According to me answer will be b and d .
But answer is given (d) , is it correct?
As you wrote " V is just range space of M ". It means that V is all the possible combinations of COLUMNS in M. Visually, the first column is just the x-axis in 3d space. So take any point on the x-axis for start. Now add a point in the directionof the second column (1,1,1). But hey, the first coordinate does not give us more freedom, since we can choose any cooridnate that we want along the x-axis. A bit more formally, the linear combinations that you can get with (1,0,0) and (1,1,1) are the same as the linear combinations of (1,0,0) and (0,1,1). This is not any plane in the world. For example, this pair does not span the vector (1,2,3). This is because the last two entries in the vector a*(1,0,0)+b*(0,1,1) would be the same: a0+b1=b.
So far the first two vectors are independent and together span a plane in R^3. Not a line, but a plane. A very specific plane. What about the third vector? If it will be independent of the first two then we we will have a basis of R^3 and any orthogonal basis of R^3 will be the answer.
However, this is not the case. The third column is just the sum of the first two columns and does not give us "more freedom". Our plane remains the same: the set {a(1,0,0)+b(1,1,1) | a,b in R}={a(1,0,0)+c(0,1,1) | a,c in R} . "Visually" this is the set of vectors in R^3 whose last two entries are the same. This is exactly the span of the vectors in (b):you can choose the first entry to be whatevery you want due to the first vector (1,0,0). Then multiply the second vector by an appropraite coefficient to get the last two (same) entries.
Answer (d) seems incorrect to me: the vector (1,0,1) is the sum of the two vectors (1,0,0)+(0,0,1) but its last two entries are different. So I can't see how can you get it via the columns of M (that all have same last two entries).
I am not sure if my answer is so visual, since the observations are more technical. I got my visual intuition for linear algebra from this great animated web: http://websites.uwlax.edu/twill/svd/action/index.html
Regarding your claim that you can choose any basis B and get any plane you want. If after my answer you still think it is correct, consider the extreme case where M is just a 3-by-3 matrix of zeroes. Thet set {Mx | x in B} would consists of only the single vector (0,0,0). You can change the base B forever and it still be correct. In fact, you can choose x to be any vector in R^3 and this would be the case. My suggestion is to first try to understand what is the column span of the given matrix (M in this case) and how the plane looks like. Only then continue to find a base, then orthogonal vectors, and finally orthonormal vectors for this base.
This is my first answer in this forum, so hope that it helps and not too long.