# How to explain that $A\vec{x}=0$ has infinite solutions when $A$ has 100 equations in 100 unknowns, and elimination reduces the 100th eq to $0=0$?

There is a problem in Gilbert Strang's Introduction to Linear Algebra that starts like this

32 Start with $$100$$ equations $$A\vec{x}=0$$ for $$100$$ unknowns $$\vec{x}=(x_1,...,x_{100}$$. Suppose elimination reduces the 100th equation to $$0=0$$, so the system is singular.

Let's call the upper triangular matrix obtained after the elimination steps $$U$$. Each column of $$U$$ is a 100d vector. The first 99 such vectors are linearly independent, and together they form a 99d "space".

The last column vector is a linear combination of the first 99 vectors. Thus, there are infinite combinations of the 100 column vectors that give the zero vector. Hence $$A\vec{x}=0$$ has infinite solutions.

My question concerns how to provide a similar interpretation from the point of view of rows of $$U$$.

Each row of $$U$$ represents a plane in 100d passing through the origin.

In 3d, $$U\vec{x}=0$$ represents three planes passing through the origin. If two of the rows of $$U$$ are linearly independent, then the two planes they represent intersect at a line passing through the origin. For the final row, it can either intersect the other two at a single point (the origin) or a line. The latter case occurs when this third row is a linear combination of the other two rows. When this happens, the normal vector to the third plane is the sum of the normal vectors to the other two planes; so all three normal vectors lie on the same plane. Intuitively, this means the three planes all pass through the same line through the origin, and are simply rotated about this "axis" line.

Now, back to 100d.

99 of the rows are linearly independent. I'm guessing they intersect on a line (though I have no idea if this is true). The last 100d vector then intersects this line.

I'm guessing that, for example, in 4d, two 3d planes intersect on a 2d plane, and three 3d planes intersect on a 1d line. Kinda hard to visualize, and I don't have the tools to investigate if this is true yet.

But I was wondering if this reasoning is true?

• Gaussian elimination (in particular backward substitution) to produce a reduced echelon form will do the trick. Dec 13, 2022 at 19:49
• @OliverDíaz Did you even read the question at all?
– xoux
Dec 13, 2022 at 20:17
• Yes I did, forward Gaussion elimination yields the last row $0$, fine. Then backwards substitution will show that there are infinitely many solutions. Dec 13, 2022 at 20:18
• @OliverDíaz That backward substitution will lead to infinitely many solutions is obvious. The bulk of my question involved a "visual", "intuitive", "geometric" reasoning about what the rows represent. The reason I even asked the question is to understand why there are infinite solutions in a manner that attains intuition, not mechanical application of an algorithm.
– xoux
Dec 13, 2022 at 22:55
• visual? its not clear what you mean, as for intuitive, I think that is much easier since your system is homogeneous in which case you know that wither there is only one solution or infinitely many. Dec 13, 2022 at 22:58

Your rough intuition at the end is right. In 100 dimensions, each row represents a 99-dimensional hyperplane. If you consider $$k$$ rows that are all linearly independent, then they will intersect in a $$(100-k)$$-dimensional hyperplane. So 2 rows would intersect in a 98-dimensional subspace, 10 rows would intersect in a 90-dimensional subspace, and 99 rows would intersect in a regular old 1-dimensional line.